Math Level I

Introduction to the SAT IITHE SAT II SUBJECT TESTS ARE CREATED and administered by the College Board and the Educational Testing Service (ETS), the two organizations responsible for producing the dreaded SAT I (which most people call the SAT). The SAT II Subject Tests are meant to complement the SAT I. Whereas the SAT I tests your critical thinking skills by asking math and verbal questions, the SAT II Subject Tests examine your knowledge of a particular subject, such as Writing, U.S. History, Physics, or Biology. The SAT I takes three hours; the Subject Tests take only one hour. In our opinion, the SAT II Subject Tests are better tests than the SAT I because they cover a definitive, easily studied topic rather than ambiguous critical thinking skills. However, just because the SAT II Subject Tests do a better job of testing your knowledge of a subject doesn¶t mean the tests are necessarily easier or demand less studying. A ³better´ test isn¶t necessarily better for you in terms of how easy it will be. The Good y y y Because SAT II Subject Tests cover specific topics, you can study for them effectively. If you don¶t know a topic in mathematics, such as how to find the slope of a line, you can easily look it up and learn it. The SAT II tests are straightforward: if you know your stuff, you¶ll do well. Often, the classes you¶ve taken in school have already prepared you for the SAT IIs. If you¶ve taken two years of algebra and a year of geometry, you¶ll have studied the topics covered by the SAT II Math IC. All you need is some refreshing and refocusing, which this book provides. In preparing for the Math, History, or Chemistry SAT II tests, you really are learning math, history, and chemistry. In other words, you are gaining valuable, interesting knowledge. If you enjoy learning, you might actually find the process of studying for an SAT II test to be worthwhile and gratifying²few can say the same about studying for the SAT I. The Bad Because SAT II Subject Tests quiz you on specific knowledge, it is much harder to ³beat´ or ³outsmart´ an SAT II test than it is to outsmart the SAT I. For the SAT I, you can use all sorts of tricks or strategies to figure out an answer. There are far fewer strategies to help you on the SAT II. To do well on the SAT II, you can¶t just rely on your natural smarts and wits. You need to study. Colleges and the SAT II Subject Tests We¶re guessing you didn¶t sign up to take the SAT II just for the sheer pleasure of it. You probably want to get into college, and know that the only reason for taking this test is that colleges want or require you to do so. Colleges care about SAT II Subject Tests for two reasons. First, the tests demonstrate your interest, knowledge, and skill in specific subjects. Second, because SAT II tests are standardized, they show how your knowledge of Math (or Biology or U.S. History) measures up to that of high school students nationwide. The grades you get in high school can¶t be compared in the same way: some high schools are more difficult than others, and students of equal ability might receive different grades, even in classes with relatively similar curriculum. When it comes down to it, colleges like the SAT IIs because the tests make the colleges¶ job easier. SAT II tests allow colleges to easily compare you to other applicants, and provide you with an excellent opportunity to shine. If you got a 93% on your Algebra final, and a student at another high school across the country got a 91%, colleges don¶t know how to compare the two grades. They don¶t know whose class was harder or whose teacher was a tougher grader. But if you get a 720 on the SAT II Math IC, and that other kid gets a 650, colleges will recognize the difference in your scores. Occasionally, colleges use SAT II tests to determine placement. For example, if you do very well on the SAT II Math IC, you might be exempted from a basic math class. It¶s worth finding out whether the colleges you¶re applying to use the SAT II tests for this purpose. Page 1 College Placement Scoring the SAT II Subject Tests There are three different versions of your SAT II score. The ³raw score´ is a simple score of how you did on the test, like the grade you might receive on a normal test in school. The ³percentile score´ compares your raw score to all the other raw scores in the country, letting you know how you did on the test in relation to your peers. The ³scaled score,´ which ranges from 200±800, compares your score to the scores received by all students who have ever taken that particular SAT II. The Raw Score You will never know your SAT II raw score, because it is not included in the score report. But you should understand how the raw score is calculated, because this knowledge can affect your strategy for approaching the test. A student¶s raw score is based solely on the number of questions that student got right, wrong, or left blank: y y y You earn 1 point for every correct answer. You lose 1/ 4 of a point for each incorrect answer. You receive zero points for each question left blank. Calculating the raw score is easy. Count the number of questions you answered correctly and the number of questions answered incorrectly. Then multiply the number of wrong answers by 1/4, and subtract this value from the number of right answers. The Percentile Score A student¶s percentile is based on the percentage of the total test-takers who received a lower raw score than he or she did. Let¶s say, for example, you had a friend named John Quincy Adams, and he received a score that placed him in the 37th percentile. This percentile score tells John that he scored better on the SAT II than 36 percent of the other students who took the same test; it also means that 63 percent of the students taking that test scored as well as or better than he did. The Scaled Score ETS takes your raw score and uses a formula to turn it into the scaled score of 200±800 that you¶ve probably heard so much about. The curve to convert raw scores to scaled scores differs from test to test. For example, a raw score of 33 on the Math IC might scale to a 600, while the same raw score of 33 on the Math IIC will scale to a 700. In fact, the scaled score can even vary between different editions of the same test. A raw score of 33 on the February 2004 Math IIC might scale to a 710, while a 33 in June of 2004 might scale to a 690. These differences in scaled scores exist to accommodate varying levels of difficulty and student performance from year to year. Which SAT II Subject Tests to Take There are three types of SAT II tests: those you must take, those you should take, and those you shouldn¶t take. y y y The SAT II tests you must take are those that are required by the colleges you are interested in. The SAT II tests you should take are tests that aren¶t required, but which you¶ll do well on, thereby impressing the colleges looking at your application. You shouldn¶t take the unrequired SAT II tests that cover a subject you don¶t feel confident about. Page Determining Which SAT II Tests are Required 2 You¶ll need to do a bit of research to find out if the colleges you¶re applying to require that you take a particular SAT II test. Call the schools you¶re interested in, look at their websites, or talk to your guidance counselor. Often, colleges require that you take the following SAT II tests: y y y The Writing SAT II test One of the two Math SAT II tests (either Math IC or Math IIC) Another SAT II in some other subject of your choice Not all colleges follow these guidelines; you should take the time to verify what tests you need to take in order to apply to the colleges that interest you. Deciding Which Math SAT II to Take Few students take both Math SAT II tests. Instead, you should choose which test to take based on several factors. y Test content. The two tests cover similar topics, but the Math IIC covers more material than the Math IC does. Level IC covers three years of college-preparatory math: two years of algebra and one year of geometry. Level IIC assumes that in addition to those three years, you have also taken a year of trigonometry and/or precalculus. Math IC Algebra Plane geometry (lines and angles, triangles, polygons, circles) Solid geometry (cubes, cylinders, cones, spheres, etc.) Coordinate geometry (in two dimensions) Trigonometry (properties and graphs of sine, cosine, and tangent functions, identities) Algebraic functions Statistics and sets (distributions, probability, permutations and combinations, groups and sets) Miscellaneous topics (logic, series, limits, complex and imaginary numbers) Math IIC (covers all areas in Math IC with some additional concepts) Algebra Plane geometry Page 3 If you have the skills to take the Level IIC test. On the IC test. Deciding If You Should Take an SAT II That Isn¶t Required There are two rules of thumb for deciding which additional test to take beyond the Writing and Math tests: 1. Try to show breadth. The following Page 4 . If history is your field. you would probably need to answer all the questions correctly to get a perfect score. Below we have included a list of the most popular SAT II tests and the average scaled score on each. it also covers the basic topics in more difficult ways than the Math IC does. require the Math IIC test. Please note that if you are planning to attend an elite school. As you choose between the two tests. vectors. you might very well get a better score than you would on the IC. Some students decide to take the Math IC because it¶s easier. keep in mind the specific colleges you¶re applying to. but some may prefer that you take the IIC. such as MIT and Cal Tech. But if you get tripped up by just one or two questions on the Math IC. and Biology. Colleges will be more impressed by a student who does fairly well on SAT II Math IIC than one who does very well on SAT II Math IC. Not only does the Math IIC cover additional topics. parametric equations) Trigonometry (cosecant. and whether or not you can get that score (or higher).Solid geometry Coordinate geometry (in two and three dimensions. however. polar coordinates. inverse functions. College choice. Colleges with a strong focus on math. in non-right triangles) Statistics and sets Miscellaneous topics y y y Question difficulty. and Chemistry will not be as impressive as good scores in more diverse subjects. Writing. Most other colleges have no such requirement. the friendly curve of the Math IIC means that if you know enough math to take the IIC. The Level IIC test is scored on a much more liberal curve: you can miss six or seven questions and still achieve a score of 800. If you feel confident that you can get a score that is above the average (50 points or more). even though they have taken a precalculus course. a strong score on the American History test will impress admissions officers far more than a bold but mediocre effort on the Physics test. taking the test will probably strengthen your college application. you would need around 20 correct answers on the IIC test and 33 on the IC test. Of course. Scoring well on similar subject tests such as Math. Biology. secant. World History. Battle of the test curves. Go with what you know. If you wanted to score a 600 on either test. you also have to know what is considered a good score. you might have to score significantly higher than the national average. Also. try a taking a practice test for each. your score will not be as impressive as you might expect. you should go for it. cotangent functions. Some students with strong math backgrounds think that they can get a marvelous score on the less difficult Math IC while their score on the IIC will only be average. We don¶t recommend this. If after all this you still can¶t decide which of the two Math SAT IIs to take. 2. such as Math. can do about it. TEST AVERAGE SCORE 590±600 590±600 580±590 570±580 580±590 655±665 590±600 605±615 635±645 Writing Literature American History World History Math IC Math IIC Biology Chemistry Physics As you decide which test to take. or you. when the material is still fresh in your mind.) After grumbling.collegeboard. History in eleventh grade.com or do some research in your school¶s guidance office. History near the end of that year.) ETS usually sets testing dates for SAT II Subject Tests in October. however. It¶s a good idea to call the schools that interest you or talk to a guidance counselor to get a more precise idea of what score you should be shooting for. you still have to register. sadly. you have to fill out some forms and pay a registration fee. go to www. it¶s best to take those after you¶ve had as much study in the area as possible. there isn¶t anything we.table is just a general guideline. we know²it¶s ridiculous that you have to pay for a test that colleges require you to take in order to make their jobs easier. November. be realistic with yourself. To check when the test you want to take is being offered. and June. Unless the colleges you¶re applying to use the SAT II for placement purposes. there is no point in taking any SAT II tests after November of your senior year. To register by mail. then you should take the SAT II U.collegeboard. But. We know. since you won¶t get your scores back from ETS until after the college application deadline has passed. (It¶s acceptable here for you to grumble about the unfairness of the world. you take U. There are two ways to go about it: online or by mail. However.com. Literature. To register online. for example. and Foreign Language SAT II tests. visit the College Board website at www. January.S.S. If. May. December. fill out and send in the Page 5 . (This rule does not apply for the Writing. When to Take an SAT II Subject Test The best time to take an SAT II Subject Test is right after you¶ve finished a year-long class in that subject. Don¶t just assume you¶re going to do great without at least taking a practice test and seeing where you stand. not every subject test is administered in each of these months. Registering for SAT II Tests To register for the SAT II test(s) of your choice. Content of SAT II Math IC The Math IC test covers a variety of topics. and scored. you¶ll still have to pay a separate registration fee for each. Introduction to SAT II Math IC The key to success on any test is simple: know your subject. or writing to: College Board SAT Program P. You¶ll know what to expect before you even enter the testing room. you might spend so much energy trying to figure out how to take the test that you¶d only get halfway through it. provides the following breakdown of coverage: Topic Algebra Plane Geometry Solid Geometry Coordinate Geometry Trigonometry Functions Statistics and Sets Miscellaneous Percent of Test 30% 20% 6% 12% 8% 12% 6% 6% Usual Number of Questions 15 10 3 6 4 6 3 3 This breakdown is accurate. the company that writes the test. but it is too broad to help you direct your studying in any meaningful way. You can also request a copy of the Bulletin by calling the College Board at (609) 7717600 (609) 771-7600 . NJ 08541-6200 You can register to take up to three SAT II tests for any given testing day. even if you decide to take three tests in one day. Box 6200 Princeton.O. We¶ve broken down the Math IC by content and format. with no preparation besides having read a textbook. and no knowledge of how you¶d even be tested. That¶s where this chapter comes in handy.forms enclosed in the Registration Bulletin. But just knowing the material isn¶t enough to guarantee a good score on SAT II Math IC²if you walked into an exam completely blind. which should be available in your high school¶s guidance office. organized. giving you a behind-the-scenes look at how your exam is written. That¶s why we created this more detailed breakdown of the test: Page 6 Topic Percent of Test Usual Number of Questions . ETS. Unfortunately. cylinders. solids by rotation Coordinate Geometry Lines and distance Graphing Conic sections (parabolas.Algebra Arithmetic Equation solving Binomials. cones. polynomials. circles Solid Geometry Solids (cubes. quadratics 30% 1±3% 18±22% 5±7% 20% 3±5% 14±18% 6% 7±9% 1±3% 12% 7±9% 1±3% 3±5% 8% 3±5% 1±3% 12% 7±9% 1±3% 15 1 10 3 10 2 8 3 4 1 6 4 1 2 4 2 1 6 4 1 Plane Geometry Lines and angles Triangles. polygons. inverse functions Graphing functions Page 7 .) Inscribed solids. etc. cosine. circles) Trigonometry Basic functions (sine. compound. tangent) Trigonometric identities Functions Basic. each question in the practice tests at the back of this book is grouped by the above categories. The instructions for the test are very simple.5 3 0. If the exact numerical value is not one of the choices. no matter their difficulty.5 Statistics and Sets Mean. Also. select the choice that best approximates this value. You can skip around while taking the test.5 1 0. allowing you to focus on each topic to whatever degree you feel necessary. For each of the following problems. decide which is the BEST of the choices given. mode Probability Permutations and combinations Group questions. median. Then fill in the corresponding oval on the answer sheet. as we explain in the next chapter. so that you can very precisely identify your weaknesses and then use this book to address them Format of SAT II Math IC SAT II Math IC is a one-hour test composed of 50 multiple-choice questions.Domain and range of functions 1±3% 1±3% 6% 1±3% 1±2% 1±2% 6% 1±2% 1±3% 1±2% 1±2% 2 2 3 1 0. The ability to skip the occasional question is helpful. and the hardest questions are last. . Have you read the directions? Have you memorized them? Good. All questions are worth the same number of points.5 0. you should memorize them so you don¶t waste time reading them on the day of the test. sets Miscellaneous Arithmetic and geometric series Logic Limits Imaginary numbers This book is organized according to these categories. the moderately difficult questions are in the middle.5 0. Now here¶s some specific information about the test¶s format: y y y The Calculator Page 8 The 50 questions progress in order of difficulty: the easiest questions come first. These points combined equal your raw score. First.Unlike the SAT I. you earn zero points. ETS converts your raw score to a scaled score according to a special curve tailored to the particular test you take. tangent Make sure you practice each of these functions on your calculator before taking the test. you earn one point. or needs to be plugged in are prohibited. make sure you have the right type of calculator. Whatever calculator you use for the test should have all the following functions: y y y Exponential powers Base-10 logarithms Sine. you lose 1/4 of a point. that¶s what the ³C´ in IC stands for. cosine. Use this table to convert your raw scores on practice tests into an approximate scaled score. We have included a generalized version of that curve in a table below. For every right answer. minicomputers. We tell you more about how to use calculators for the test in the next chapter. Scoring SAT II Math IC Scoring on the SAT II Math IC is the same as the scoring for all other SAT II tests. Some questions on the test are specifically designed to test your calculator-using skills. or any machine that prints. Virtually any calculator are may be used during the test. For every wrong answer. Laptops. For every answer left blank. the Math IC test demands the use of a calculator. Average Raw Score 50 49 48 47 46 45 44 43 42 41 Scaled Score 800 780 770 760 740 730 720 710 700 690 Average Raw Score 18±19 17 16 15 14 13 12 11 10 9 Scaled Score 480 470 460 450 440 430 430 420 410 400 Page 9 . in which a calculator is permitted but not essential to the test. including programmable and graphing calculators. It is therefore wise to learn all the essentials about calculators before taking SAT II Math IC. makes noise. In fact. 40 39 38 37 36 35 34 33 32 31 30 29 28 27 26 25 24 23 22 21 680 670 660 650 640 630 610 600 590 580 570 560 550 550 540 530 520 510 510 500 8 7 6 5 4 3 2 1 0 ±1 ±2 ±3 ±4 ±5 ±6 ±7 ±8 ±9 ±10 ±11 390 380 370 370 360 350 340 330 330 320 310 300 300 290 280 270 260 260 250 240 Page 10 . But it is amazing how a timed test can warp and mangle common sense. and left 7 blank These sample scores suggest that when taking the test. because if you automatically assume you won¶t be able to answer many of the questions. So we¶ve organized a few of the basic (and not-so-basic) rules and tips for test-taking that will best enable you to get those right answers quickly. nothing else.20 490 ±12 230 As you can see. On a 50-question test. The first type results from sheer overconfidence. this curve is not very forgiving. and left 3 blank 650 if you answered 38 right. If you answered a question correctly. For example. The computer that scores SAT II tests is unmerciful. the key to doing well on SAT II Math IC is to follow a strategy that ensures you will see and answer all the questions you can. you shouldn¶t imagine your score plummeting with every question you can¶t confidently answer. These rules are so obvious that we hesitate to even call them ³strategies. no matter how comfortable you are with the material. and left 1 blank 740 if you answered 46 right. you make yourself vulnerable to misinterpreting questions. the computer will mark that question as Page 11 Be Careful Gridding In Your Answers . while intelligently guessing on those slightly fuzzier questions. Even if you don¶t feel confident about the material. There¶s nothing worse than realizing you lost points due to sloppy mistakes. you can miss a bunch of questions on the Math IIC and still get the same score you would receive on the Math IC if you missed just one. not the thoughts behind them. As you take the test. Avoid Carelessness There are two types of carelessness. 0 wrong. But all is not hopeless on the SAT II Math IC. Don¶t get unnecessarily wound up if you run into a difficult question. If you speed through the test without a second glance. and you might find you know more (and get a better score) then you thought you would. So we offer the following list. 8 wrong. but somehow made a mistake in marking your answer grid. you could score: y y y y y y 780 if you answered 49 right. overlooking answer choices. 0 wrong. Believe it or not. and making computational mistakes. 4 wrong. NOT A PERSON. you can use all this to your advantage. you¶ll get a point. and left 7 blank 600 if you answered 35 right. stay on track and use our techniques for test-taking. and left 4 blank 700 if you answered 43 right. We discuss these strategies in the next chapter. Basic Rules of SAT II Test-Taking There are some rules of strategy that apply to all SAT II tests. you¶ll give up at the first sign of difficulty and sabatoge your score. Strategies for SAT II Math IC A MACHINE. make a conscious effort to approach it calmly and methodically. Reiterating what we said earlier.´ Some of these rules will seem more like common sense to you than anything else. and left 3 blank 650 if you answered 39 right. We don¶t disagree. both of which will cost you points. Then there¶s lack of confidence²a defeatist attitude is your worst enemy when taking the SAT IIs. ETS only wants right answers. a raw score of 41 on the Math IIC test receives an equivalent scaled score as a raw score of 49 on the Math IC test. The tabulating machine sees only the filled-in ovals on your answer sheet. 8 wrong. Getting just one question wrong will lower your score by 20 points. So whether you knew the right answer because you¶re a math genius or because you took a lucky guess. 4 wrong. WILL SCORE your Math IC SAT II test. . the test booklet is yours to write all over. . exhaustively checking their work and wasting time that they¶ll need for the tricky second .´ Page 12   now What¶s in the Reference Area At the beginning of SAT II Math IC. . . and the answer to question 7 in row 6. Nobody will look at or reward your work. but put the answer to question 6 in row 5. Mark up graphs or charts as necessary. a helpful little gnome whispers to you. easy question. don¶t neglect to memorize and understand the formulas because you have the reference area as a crutch. A. and writing can often help clarify things. Number 24.´ Would the gnome¶s statement affect the way you approach the two problems? Yes. etc. After your teacher hands out the test. THE FOLLOWING INFORMATION IS FOR YOUR REFERENCE IN ANSWERING SOME OF THE QUESTIONS IN THIS TEST. you¶ll save time. but other people don¶t. the second is much harder. You need to be able to follow and understand your work. Draw diagrams or write out equations to help you think. both to come up with an answer and to check your work to make sure you didn¶t make an error somewhere along the way. it gets ugly. You must find a balance between speed and accuracy. Number 25. You will probably have to spend much more time on a ³much harder´ question. . . but doing pristine work. and before you set to work. B. Write All Over Your Test Booklet . it seems likely that you should be able to answer it quickly and with little or no agonized second-guessing. or more work than necessary. In our opinion. Doing math scratchwork can definitely help you avoid careless errors. . we must qualify our advice. Volume of a right circular cone with radius r and height h: Lateral area of a right circular cone with circumference of the base c and slant height l: Volume of a sphere with radius r: Surface area of a sphere with radius r: Volume of a pyramid with base area B and height h: You should know all these formulas without needing the reference area. If you skipped question 5. there is a reference area that provides you with basic geometric formulas and information. see the reference area as a hint to you about what formulas are likely to be needed on the test. the best way to ensure that you¶re being careful is to talk silently to yourself. say to yourself: ³Number 23. We think you should fill out the answer sheet whatever way feels most natural to you. Cross out answers that can¶t be right. If you know those formulas without having to flip back to the reference area. just make sure you¶re careful while doing it. allowing you to work more quickly with fewer mistakes. The Importance of the Order of Difficulty Imagine that you are taking a test that consists of two questions. so don¶t write it out as if you¶re being judged. thereby throwing off your answers for an entire section . Basically. Some suggest that you do one question and then fill in the corresponding bubble. ³The first problem is very simple. Instead. As you figure out an answer in the test booklet and transfer it over to the answer sheet. Some test-prep books advise that you fill in your answer sheet five questions at a time rather than one at a time. For a ³very simple´ question. can be more time-consuming than it¶s worth. What about all the other students who didn¶t hear the gnome? They might labor over the first. which puts you one step ahead. .wrong. E. But Remember that the SAT Rewards Answers. Not Work That said. So you quickly multiply 6 and 8 to get 42 and then multiply 42 by 3 to get 126. 6 8 equals 48. Let¶s do it: Area of a square = s2. and devote more time to the harder questions appearing later. Area of a circle = r2. Then. you should be more cautious. These different approaches to answering questions vary in the amount of time they take. the test tips you off about when to take a few extra seconds to make sure you haven¶t been fooled by an answer that only seems right. since the equation for the circle will square the 4 and multiply it by . so the circle must be bigger. With difficult questions. you might be able to answer the question quickly. not 42. The tricky answers seem right because they are actually the answers you would get if you were to make a mathematical or logical mistake while working on the problem. But a faster approach would have been to draw a quick to-scale diagram with the square and circle superimposed. since they have no idea that the problem is so difficult. 16 is obviously bigger than 16. There is a reason most people get these questions wrong: not only are they more difficult. Math Questions and Time There are often several ways to answer a Math IC question. Page An even quicker way would have been to understand the equations for the area of a square and a circle so well that it was obvious that the circle was bigger. they might not check their work or be wary of traps. The moral here is you should spend less time on the simpler questions that appear early in the test. while the elegant method of relying on an intuitive understanding of conceptual knowledge takes the least amount of time. you should learn that just because the answer you arrived at is among the answers does not mean you definitely have it right. For example. That worked nicely. you can set up and solve an equation. and the area of this circle must therefore be 42 = 16 . when those other students do get to the second problem. making the correct answer 144. it¶s as if you have that helpful little gnome sitting next to you for the entire test. You look down at the answers. intuitively. and. You can use trial and error. for some questions. What does this mean to you? It means that when you are going through the test. The SAT is designed to punish those who make careless errors. 13 ¡ nowing When to Be Wary Most students answer the easy Math IC questions correctly. . you can often trust your first instincts on an easy question. Trial and error generally takes the longest. the following problem: Which has a greater area.problem. and elegantly. if you can just spot how to do it. Only some students get moderate questions right. however. let¶s say you¶re flying through the test and have to multiply 6 8 3. and there¶s 126! You mark it down as your answer and you get the question wrong. full of enticing wrong answers that seem as if they must be correct. a square with sides measuring 4 cm or a circle with a radius of the same length? The most obvious way to solve this problem is simply to plug 4 into the formula for the area of a square and area of a circle. quickly check your work again. Take. containing more sophisticated vocabulary or mathematical concepts. they are also often tricky. Because Math IC questions are ordered by difficulty. Don¶t be one of them. From this example. for example. Very few students get difficult questions right. so the area of this square = 42 = 16. whereas the equation for the square will only square the 4. But because the SAT orders its questions by difficulty. After you get an answer. According to ETS. with examples. and quickest. above all. a shortcut requires using your existing knowledge to spot a faster way to answer the question. calculator-neutral. Making such choices comes down to practice. Making Your Calculator Work for You As we¶ve already mentioned. about 60 percent of the test falls under the calculator-neutral and -friendly categories. Here¶s an example: If f(x) = (A) (B) (C) (D) . The other 20 questions are calculator-unfriendly and useless. you know that there will probably be a shortcut for all those questions that give you the dimensions of two shapes and ask you to compare them. But at the same time. Whether or not the ability to find accurate shortcuts is an actual measure of mathematical prowess is not for us to say (though we can think of arguments on either side). That is. logarithms. From the problem above. A frantic test-taker might compulsively work out the equations every time. When we use the term math shortcut. We can guarantee that you¶re won¶t find a shortcut for a problem unless you know how to work it out the long way.73 ±16.55 Page 14 . it might end up taking longer than the typical route. But if you are a little calmer. Shortcuts Are Really Math Intuition We¶ve told you all about shortcuts.32 ±16. you can teach yourself to recognize when a question might contain a shortcut. but now we¶re going to give you some advice that might seem strange: you shouldn¶t go into every question searching for a shortcut. having an awareness that those other routes are out there. To some extent. know how to use it intelligently. calculator-unfriendly. And. the calculator is a very important part of the Math IC test. It is this issue of time that separates the students who do terrifically on the math section and those who merely do well. we¶re really referring to your math intuition. Computations that you can¶t do easily in your head are prime candidates. Calculator-Friendly Questions A calculator is extremely helpful and often necessary to solve calculator-friendly questions. then what is f(3. calculators are useful or necessary on 30 of the 50 questions on SAT II Math IC. you have a chance to find the shortcuts you need. but the ability to find those shortcuts absolutely matters on this test. you might miss the possibility that a shortcut exists. Here¶s a breakdown of each of the four types. We cover all these topics in this book. such as choosing to draw a diagram instead of working out the equation.While you may be a math whiz and just know the answer. as with the example above. and calculator-useless.4)? ±18. a quicker route is not necessarily a less accurate one. You need to have the right kind of calculator.28 ±13. solution. and. or trigonometric functions will most likely need a calculator. be familiar with its operations. don¶t worry. you can see that drawing a diagram is the best. The trick is to be able to identify the different types of questions when presented with them on the test. After all. If you go into each question knowing there might be a shortcut and keep your mind open. If you¶re not certain about the math discussed in the examples. The fact that we advocate using shortcuts doesn¶t mean you shouldn¶t focus on learning how to work out problems. and basic mathematical ability. you can learn to look for a quicker route. There are four types of questions on the test: those that are calculator-friendly. if you¶re so frantic about calculating out the right answer. If you have to search and search for a shortcut. Problems demanding exact values for exponents. The value of time-saving strategies is obvious: less time spent on some questions allows you to devote more time to difficult problems. (E) ±8.6467 . then you might not want to risk the possibility of making a mental math mistake and should choose the first method.4. So you could type in log8 (43 23) on your trusty calculator and find that x = 3. These types of problems often have built-in shortcuts²if you know and understand the principle being tested. it isn¶t a good idea. As you will learn in the Functions chapter. Calculator-Unfriendly Questions While it¶s possible to answer calculator-unfriendly questions using a calculator.7598 If you didn¶t take a moment to think about this problem. But if you¶re more prone to error when working with a calculator. A calculator is useful for these types of problems. we need to find the exponent x. Calculator-Neutral Questions You have two choices when faced with a calculator-neutral question. thinking a step further. ±16. but it¶s probably just as quick and easy to work the problem out by hand. that 23 = 8 and 43 = 64 = 82.4.42 This is a simple function question in which you are asked to evaluate f(x) at the value 3. you should think of logarithms. subtract twice the square of 3. then logab = x. From the definition of logarithms. you could recognize that 2 and 4 are both factors of 8. all you need to do is take the square root of 3. But unless you know the square root and square of 3. Put together. and. If 8x = 4 3 (A) (B) (C) (D) (E) 2 3 . If you feel quite comfortable with your calculator. you might just rush into it wielding your calculator. what is the value of x? 2 3 5 7 8 When you see the variable x as a power. Or. Here¶s a problem that you could solve much more quickly and effectively without the use of a calculator: (A) (B) (C) (D) (E) . squaring them each and then adding them Page 15 . calculating the cosine and sine functions. These two processes take about the same amount of time. you can bypass potentially tedious computation with a few simple calculations.3261 . this problem is extremely difficult to answer without a calculator. such that 8x = 43 23. We come to the same answer that x = 3 and that B is the right answer. so choosing one over the other is more a matter of personal preference than one of strategy.4.9238 .4 for the variable x and carry out the operations in the function.5 . But with a calculator. You get answer choice C. we know that if given an equation of the form ax = b. and then add 5. A logarithm is the power to which you must raise a given number to equal another number.4 off the top of your head (which most test-takers wouldn¶t). so in this case.28. 43 23 = 82 8 = 83. then you should choose the second method. all you have to do to solve this problem is plug in 3. For the most part. In our case. B is correct.182 4. You should be able to easily identify problems that can¶t be solved with a calculator. For example. So. first take a brief look at each question and understand exactly what it¶s asking you to do. To start. then what is y? A trigger-happy calculator user might immediately plug in 3 for x. the expression {cos2(3 63°) + sin2(3 63°)} 4/2 simplifies to 14 /2 = 1/2 = . A calculator would be of no use here. But take a closer look: cos2(3 63°) + sin2(3 63°) is a trigonometric identity. B is correct. problems involving algebraic manipulation or problems lacking actual numerical values would fall under this category. it¶s a Pythagorean identity: sin2q + cos2q = 1 for any angle q. Just because you¶ve got an awesome shiny hammer doesn¶t mean you should try to use it to pound in thumbtacks.together. etc. To solve this problem. a = x + y and b = 1. the answers for these questions will be variables rather than numbers. Using your calculator to try to answer every question on the test would be just as unhelpful.5. (x + y ± 1)(x + y + 1) = (x + y)2 ± 1. y) is a point on the graph of f(x) = (A) (B) (C) (D) (E) ±3 ±1. As a result. you wouldn¶t be able to use your calculator on calculator-useless problems. factor 11 out of the denominator: Then. not produce a specific value.87 . Calculator-Useless Questions Even if you wanted to. what if you came upon the question: If (3.45 0 . it asks you how to find the product of two polynomials²and requires knowledge of algebraic principles rather than calculator acumen. Don¶t Immediately Use Your Calculator The fact that the test contains all four of these question types means that you shouldn¶t get triggerhappy with your calculator. Take a look at the following example: (x + y ± 1)(x + y + 1) = (A) (B) (C) (D) (E) (x + y) 2 (x + y) 2 ± 1 x2 ± y2 x2 + x ± y + y2 + 1 x2 + y2 + 1 This question tests you on an algebraic topic²that is. factor the numerator to its simplest form: Page 16 . You¶re asked to manipulate variables. That short pause will save you a great deal of time later on. Quite often. you need to notice that the two polynomials are in the format of a Difference of Two Squares: (a + b)(a ± b) = a2 ± b2. Instead of reaching instinctively for your calculator. More specifically. But the student who takes a moment to think about the problem will probably see that the calculation would be much simpler if the function was simplified first. won¶t be equal to zero (answer C). Working Backward: The Process of Elimination If you run into difficulty while trying to solve a multiple-choice problem. then move on. hidden among five answer choices. you could build the equations: Then. At this point you could shift to the calculator and calculate f(x) = (3 ± 1) » 11 = 2/ 11 = . go ahead. some of which have arms and some of which do not. 2. especially if you work strategically. you would see that you don¶t even have to work out this final calculation. So if you can¶t solve the problem directly. there is a standard procedure that you should use to approach all of them. If the room contains 5 more armchairs than chairs without arms. however. Take the following example: A classroom contains 31 chairs. 1. Read the question without looking at the answers. Do not look at the answers unless you decide that using the process of elimination is the best way to go.182. which is answer D. Approaching Math IC Questions Though there are four types of questions on the Math IC. 2»11 can¶t work out to any answer other than D. you might want to try the process of elimination. Determine what the question is asking and come to some conclusion about how to solve it. the answer is right in front of you. Not only can this process help you when you can¶t figure out a question. how many armchairs does it contain? (A) (B) (C) (D) (E) 10 13 16 18 21 Given this question. you might be able to plug each answer into the question to see which one works. there are times when it can actually be faster than setting up an equation. If you were very comfortable with math. since y = x ± 5 you can make the equation: Page 17 . test it quickly to make sure it¶s correct. Once you¶ve derived an answer.The (x ± 4) cancels out. since you know that 2»11 isn¶t a negative number (like answers A and B). and the function becomes f(x) = (x ± 1) » 11. Once you¶ve decided on an answer. If you think you can solve the problem. For every question. only then see if your answer matches one of the choices. and also won¶t be greater than 1 (answer E). 3. But because the total number of chairs is too few. working backward will often be more difficult than actually working out the problem. (Of course. if 16 is too big. then which of the following must be odd? (A) (B) (C) (D) (E) p+q p±q p2 + q2 p2 q2 p + q2 It might be hard to conceptualize how the two variables in this problem interact. with hard fractions or radicals. 18 . There¶s our answer. If the answer choices are complicated. working backward and plugging in is not always the best method. when multiplied. you will need to build up a sense of when working backward can most help you. So our strategy is in place. as two odd numbers. For example. If we then plug in D. plugging in might prove so complex that it¶s a waste of time. But what if you chose two odd numbers. plugging in the answers takes less time. Here¶s a good rule of thumb: Work backward when the question describes an equation of some sort and the answer choices are all simple numbers. and just seems easier in general.There are 18 armchairs in the classroom. We needed the total number of chairs to equal 31. you have to remember to keep the substitution consistent. If we have 16 armchairs. we can eliminate A and B along with C.) Substituting numbers can help you transform problems from the abstract to the concrete. to represent the two variables? You get: (A) (B) (C) (D) (E) p+q=5+3=8 p±q=5±3=2 p 2 + q 2 = 25 + 9 = 34 p2 q 2 = 25 2 9 = 225 p + q = 5 + 9 = 14 Page The answer has to be D. Now. If you¶re using a 5 to represent p. In this instance. the answer choices with smaller numbers of armchairs.p2 q2 since it multiplies to 225. If the answer choices contain variables. Now let¶s work it out. take the question: If p and q are odd integers. This is a smart strategic move because if we plug in 16 and discover that it is too small a number to satisfy the equation. then we would have 11 normal chairs and the room would contain 27 total chairs. we have 13 normal chairs and 31 total chairs. Choose numbers that are easy to work with and that fit the definitions provided by the question. you could have answered this question without any work at all. This approach of building and working out the equations will produce the right answer. you have to strategically decide on numbers to substitute into the question to take the place of variables. For some questions it won¶t be possible to work backward at all. except the numbers you plug into the equation aren¶t in the answer choices. let¶s choose the one in the middle: C 16. 18. we can eliminate D and E along with C. always result in an odd number. Instead. so clearly C is not the right answer. we can also eliminate A and B. However. let¶s say 5 and 3. but it takes a long time! What if you strategically plugged in the answers instead? Since the numbers ascend in value. Alternatively. don¶t suddenly start using 3. For the test. Substituting Numbers Substituting numbers is a lot like working backward. of course! And that¶s precisely the situation you¶re in when you blindly guess the answer on any SAT II Math IC question: you have a 1 in 5 chance of getting the question right. and now have four choices from which to choose. Eight wrong answers gets you ±2 raw points (8 ± 1/4 points). the odds of guessing turn in your favor: you become more likely to gain points than to lose points. you¶ll get one point. Now is it worth it to guess? Yes. Some questions will stump you completely. Answering 50 math questions in 60 minutes is not the easiest of tasks.Guessing and the Math IC Should you guess on SAT II Math IC? We¶ll answer this question by posing a question of our own: G. you should be able to at least look at every single question on the test. Note that we said ³look at´ every question. Without telling you. you should definitely guess Pacing: The Key to Scoring Well As we said earlier. Others might demand so much of your time that answering them becomes more trouble than it¶s worth. Educated Guessing But suppose you¶re faced with this question: If x + 2x = 6. Metry is holding five cards. And that¶s exactly what ETS wants. O. numbered 1±5. y y Two right answers earns you 2 raw points. what is the probability that you will choose the correct card? One out of 5. net you a total of 0 points. probability says you¶ll get two questions right and eight questions wrong. we didn¶t say ³answer. If you were to guess on 10 questions. you¶ll lose a total of 3/4 of a point: 1 ± 3/4 = 1/4. the questions on the SAT II Math IC Test are organized from least to most difficult. You can eliminate ³0´ as a possible answer. and realize that 0 multiplied by any number equals 0. Those ten answers. what is the value of x? (A) (B) (C) (D) (E) ±2 2 3 0 1 Let¶s say you have no idea how to solve this problem. he has selected one of the numbers as the ³correct´ card.´ It is unlikely that you will be able to answer every question on the test. If you plug that into the equation. If you pick a single card. While taking five minutes to solve a particularly difficult question might strike you as a moral victory when you¶re taking the test. or 1»5. but if you learn how to pace yourself. you could have used that same time to Page 19 . For that one correct answer. and for the three incorrect answers. But you look at the answer choices. They designed the test to make blind guessing pointless. cannot add up to 6. with the basic material covered near the beginning and the advanced topics at the end. If you can eliminate even one answer. putting yourself in the position of having to leave blank those questions near the end of the test that you could have answered if only you had more time. Probability states that if you are guessing between four choices you will get one question right for every three you get wrong. Make sure you don¶t spend too much time on the easiest questions. therefore. The rule for guessing on the Math IC test is simple: if you can eliminate even one answer-choice on a question. There are a few simple rules that will make pacing yourself much easier. and left 1 blank 740 if you answered 46 right. and left 3 blank 600 if you answered 35 right. and make an educated guess on every question for which you can quickly eliminate at least one answer choice. right?). you can get 39 questions right. or talk to your guidance counselor. and left 3 blank 650 if you answered 39 right. Mark the question in some way to indicate it is very difficult. integrating your new knowledge of the test and how to take it without overwhelming yourself. you will do better if you learn to skip. Then take a look at the chart we showed you earlier. By improving your score in manageable increments. and you set your target at about 650. Take that number and set your target score above it (you want to be above average. you can slowly work up to your top speed. give yourself a cookie and take a break for the day. 4 wrong. If you meet your new target score again. but be realistic: consider how much you know about math and how well you usually do on SAT-type tests. Answer every question for which you know the answer. 8 wrong. and come back to it later. you have little chance of making an educated guess. or decreasing your pace if you find that you¶re rushing and making careless mistakes. what score do you want? Obviously. move on. 8 wrong. your score will certainly go up. According to the chart. repeat the process. If you know all these numbers going into the test. increasing your speed if you find that you aren¶t getting to answer all the questions you need to. Instead of getting bogged down on individual questions. Skip questions in which the question and answers refer to concepts completely foreign to you. get 8 wrong. you should strive for the best score possible. you should view reaching your target score as a clue that you can do better than that score: set a new target 50-100 points above your original. and still achieve your target score. the very difficult questions either that you can¶t answer or that will take an extremely long time to solve. You should also consider what exactly defines a good score at the colleges you¶re applying to: is it a 620? A 680? Talk to their admissions offices. But just because you hit your target score doesn¶t mean you should stop working altogether. You should also find out the average scores of students already at the schools you want to attend. So. If you reach your target score during preparation. and leave for later. letting you choose which questions you will and will not answer.answer six other questions that would have vastly increased your score. If you can handle working just a little faster without becoming careless and losing points. y y y Don¶t get bogged down on one single question. If you find yourself wasting time on a question. and left 7 blank Page So let¶s say the average score for SAT II Math IC for the school you want to attend is a 600. Remember to skip that line on your answer sheet! Setting a Target Score You can make the job of pacing yourself much easier if you go into the test knowing how many questions you have to answer correctly in order to earn the score you want. do a little research in college guidebooks. 0 wrong. If you look at the question and answers and have no idea what topics they cover. you can pace yourself accordingly. Return to this type of question only if you have answered everything else. and work to pick up your pace a little bit and skip fewer questions. 20 . rather than running out of time before reaching the end of the test. You¶ll get: y y y y y 780 if you answered 49 right. and left 4 blank 700 if you answered 43 right. In fact. circle it. By perfecting your pacing on practice tests. 0 wrong. you can make sure that you will see every question on the test. You should use practice tests to teach yourself the proper pace. leave 3 questions blank. and you¶ll need to be able to apply these fundamentals even when answering the most sophisticated questions. we should find out the proper way to evaluate the expression parentheses. a single expression would take on a vast array of values.Math IC Fundamentals ONLY A FEW QUESTIONS (2 to 5 percent) will directly test basic math. Let¶s work through a few examples to see how order of operations and PEMDAS work. the first operation we carry out is exponentiation: . learn it! Order of Operations The order of operations is one of the most instrumental and basic principles of arithmetic. we have PEMDAS²an acronym for determining the correct order of operations in any expression. If there¶s something you don¶t know. In order to ensure that all expressions have a single correct value. But knowledge of basic math is crucial²almost all of the test¶s 50 questions assume in-depth understanding of it. It refers to the order in which you must perform the various operations in a given mathematical expression. since those two operations are the equivalent of raising a base to the 1»2 and 1»3 power. If operations in an expression could be performed in any random order. there are a number of possible evaluations of this expression. Multiplication and Division: perform multiplication and division. PEMDAS stands for: y y y y Parentheses: first. we do all the necessary multiplication and division: Lastly. you¶ll fly right through it. If you know it all. A set of parentheses supercedes any other operation. we perform the required addition and subtraction. Our final answer is: Page 21 . Either way. Exponents include square roots and cube roots. respectively. perform the operations in the innermost parentheses. Exponents: raise any required bases to the prescribed exponent. depending on the order in which we perform the required operations. it can¶t hurt to thumb through this chapter. You probably know some of the Math IC Fundamentals like the back of your hand. Since nothing is enclosed in Next. For example: Evaluate the expression One student might perform the operations from left to right: Another student might choose to add before executing the multiplication or division: As you can see. Addition and Subtraction: perform these operations last. while others may need a refresher. First. .Here¶s another example. One additional note is important for the division step in the order of operations. Instead. To work out this expression. which is a bit trickier. you can type the whole expression into your calculator. In terms of math. first execute the operations within the innermost set of parentheses: Next. Expressions under a radical are special exceptions because they are really an expression within parentheses that has been raised to a fractional power. 2. multiply: Finally. Try it on your own. you must be familiar with how your calculator works. you must evaluate the numerator and the denominator separately before you divide the numerator by the denominator. but can cause careless errors. add: Now that the operations under the radical have been resolved. The fraction bar is the equivalent of placing a set of parentheses around the whole numerator and another for the whole denominator. resolve the operations under the square root. You can¶t enter fractions and exponents into your calculator the way they appear on paper. If you want to type full expressions into your graphing calculator. When the division symbol is replaced by a fraction bar (i. But wait. Order of Operations and Your Calculator There are two ways to deal with the order of operations while using a calculator: 1. . which is symbolized by and is also called a radical. This is a slow but accurate process. you may be thinking to yourself. If you have a graphing calculator. perform the required exponentiation: Then. Work out operations one by one on your calculator while keeping track of the entire equation on paper. we can take the square root. First.e. Practice with the following expression: Page 22 . you have to be sure to recognize and preserve the order of operations. The radical effectively acts as a large set of parentheses. and then compare your results to the explanation that follows: Evaluate . the expression includes a fraction). This method will be faster. so the rules of PEMDAS still apply. I thought we were supposed to do everything within a parentheses before performing exponentiation. . Rational Numbers. . . . no number can be both. the most important ones to understand are probably integers and real numbers. 3. The set of all whole numbers except zero {1. any number that can be expressed in the form m»n . 1. . where m and n are integers. . 3. Only integers can be even or odd. ±4.}. Real Numbers.If you enter this into a graphing calculator. ±2. including zero {0. Natural Numbers. . Even and Odd Numbers Even numbers are those numbers that are divisible by two with no remainder. ±3. . . . . . 1. it should look like this: Numbers Before you take the Math IC. Zero. 2. 1. . 5. Integers. Irrational Numbers. 2. . . The sets of irrational numbers and rational numbers are mutually exclusive. The set of counting numbers. . . They can be spotted in nearly every question on the test and will be explicitly mentioned at times. . Examples include . .01001000100001000001 . 0. The set of real numbers includes all rational and irrational numbers. Imaginary Numbers. ADDITION: . Any given number must be either rational or irrational. . including zero. . you should know the common types of numbers. 2. Operations of Odd and Even Numbers even + even = even odd + odd = even Page 23 There are a few basic rules regarding the operations of odd and even numbers that you should know well. meaning decimals and fractions are not included. where n is an integer. 6. Odd numbers are those numbers not evenly divisible by two. The set of all numbers that cannot be expressed as a quotient of integers. . If you grasp the principles behind the two types of signed numbers. however. 4. Of these types. . ±3. . Every number on the number line. these rules should all come easily. 2. The set of even numbers and the set of odd numbers are mutually exclusive. ±1.}. y y y y y Whole Numbers. 1. and odd numbers are not. 3.}. integers and real numbers will appear far more often than any of the other number types. Odd numbers are the numbers that can be written in the form 2n + 1. See the ³Miscellaneous Math´ chapter later in this book. . . ±6. 5. A more rigorous definition of even and odd numbers appears below: Even numbers are numbers that can be written in the form 2n. . y y On the Math IC. The set of rational numbers includes all integers and all fractions that can be created using integers in the numerator and denominator. ±2. is an integer and thus a member of the set. . It may come in handy. . ±1. ±5. when you need to represent an even or odd number with a variable. 0. Fractions and decimals are not included {. 4. This definition is nothing more than a technical repetition of the fact that even numbers are divisible by two. where n is an integer. That is. though. The set of all positive and negative whole numbers. The set of all numbers that can be expressed as a quotient of integers. . 3. and the absolute value of a negative number is the opposite of that number. you could think of it as the positive ³version´ of every number. Or. Operations of Positive and Negative Numbers The following rules define how positive and negative numbers operate under various operations. the answer is often ambiguous. ADDITION AND SUBTRACTION: When adding and subtracting negative numbers.even + odd = odd SUBTRACTION: even ± even = even odd ± odd = even even ± odd = odd MULTIPLICATION AND DIVISION: even odd even even = even odd = odd odd = even Positive and Negative Numbers Positive and negative numbers are governed by rules similar to those that have to do with even and odd numbers. For example: Subtracting a negative number is the same as adding its opposite. For example: MULTIPLICATION: positive negative positive positive = positive negative = positive negative = negative positive = positive negative = positive negative = negative DIVISION: positive negative positive The rules for multiplication and division are exactly the same since any division operation can be written as a form of multiplication: a b = a/b = a 1/b. The absolute value of x is symbolized by |x|. it helps to remember the following: Adding a negative number is the same as subtracting its opposite. The absolute value of a positive number is that same number. Take a look at the following equation: . As you will see. Absolute Value The absolute value of a number is the distance on a number line between that number and zero. The number zero is neither positive nor negative. First. for their quick definitions: Positive numbers are numbers that are greater than zero. Negative numbers are numbers that are less than zero. Page 24 Solving an equation with an absolute value in it can be particularly tricky. beginning with 1 and the number you¶re factoring. in increasing order. the next factor of 24. then b is a factor of a. One and 24 are both factors of 24. We¶ll factor 24 in this example. All prime numbers are positive (because every negative number has ±1 as a factor in addition to 1 and itself). Is 91 divisible by 5? No. 91 is not prime. Sometimes it is necessary or helpful to factor an integer completely. 4. y y y y Therefore. Such an effort would take an incredible amount of time. and 10 is not divisible by 3. are factors of 12. you should estimate the square root of the number: . 91 does not end with 0 or 5. Is 91 divisible by 3? No. all you need to do is estimate the square root of the number. Next. 3. In either case. had already been included in a pair of factors. are: To determine whether a number is prime. after you found that 4 was a factor of 24 and 5 was not. to decide whether a number is prime. we¶ll deal more with absolute values in equations later on in the Algebra chapter. and 6. Keep this in mind. Factors A factor is an integer that divides another integer evenly. It¶s possible that the test will directly require this skill or will make use of it in a more complicated question. you would see that 6. Prime Numbers A prime number is a number whose only factors are 1 and itself. to see if 91 is prime. The numbers 3. The first few primes. and you have only an hour for the Math IC. Factorization To find all the factors of a number. For example. If a /b is an integer. Here are the factor pairs we find for 24: y y y y 1 and 24 (1 2 and 12 (2 3 and 8 (3 4 and 6 (4 24 = 24) 12 = 24) 8 = 24) 6 = 24) You know you¶ve found all the factors of a number when the next first factor exceeds its corresponding second factor. all the factors have been found. Is 91 divisible by 7? Yes! 91 7 = 13. Thus. then check all the prime numbers that fall below your estimate. it¶s something you should know how to do. all prime numbers besides 2 are odd. Page 25 Is 91 divisible by 2? No. try every integer greater than 1 in increasing order. for example. Furthermore. This means you need to find all the factors of that integer. 9 + 1 = 10. 5 and 7. it does not end with an even number. Instead. . For example. write them down in pairs. you shouldn¶t check whether the number is divisible by every number less than itself. Now you should test 91 for divisibility by the prime numbers smaller than 10: 2.We can simplify the equation in order to isolate |x|: Knowing that |x| = 2 means that x = 2 and x = ±2 are both possible solutions to the problem. and then determined the prime factorization from there. We could have first resolved 36 into 6 6. for example? First. Just for practice. What is the greatest common factor of 18 and 24. Since the only factors of 41 are 1 and 41. It can be helpful to think of prime factorization in the form of a tree: As you may already have noticed.Prime Factorization Another form of factorization is called prime factorization. So don¶t worry²you can¶t screw up. That is. In this case.´ or intersection. Finding the GCF of two numbers is especially useful in certain applications. the GCF is the ³overlap. 41 is a prime number. such as manipulating fractions (we explain why later in this section). find the prime factorizations for 45 and 41. That is to say. divide it and all its factors until every remaining integer is prime. as long as you do your arithmetic correctly. The prime factorization of an integer is the listing of the prime numbers whose product is that number. It is therefore its own prime factorization. let¶s find the prime factorization of 36. In order to find the GCF of two numbers. their prime factorizations: The greatest common factor is the greatest integer that can be written as a product of common prime factors. Here¶s another example: Page First: 26 What is the GCF of 96 and 144? . No matter which path you take. there is more than one way to find the prime factorization of a number. of the two prime factorizations. you will always get the same result. Greatest Common Factor The greatest common factor (GCF) of two numbers is the greatest factor that they have in common. both prime factorizations contain 2 3 = 6. To find the prime factorization of a number. we must first produce their prime factorizations. This group of prime numbers is the prime factorization of the original integer. This is their GCF. for example. As an example. The GCF is 22 5 = 20. which is their GCF. are all multiples of 9.So. 5. because 1 n = n. Multiples A multiple is an integer that can be evenly divided by another integer. 2. useful when manipulating fractions: For example. 3. find the GCF of the following pairs of integers: 1. This doesn¶t mean. 30 = 2 3 5. and 18. Alternatively. 13 = 1 13. If c /d is an integer. The GCF is 7.e. the product of the prime factors that they share is 24 3 = 48. that each number is itself prime. The numbers 8 and 15 are relatively prime because they have no common primes in their prime factorizations (8 = 2 2 2 and 15 = 3 5). 15 = 3 5.. 12 and 15 30 and 45 13 and 72 14 and 49 100 and 80 Compare your answers to the solutions: 1. The numbers 45. note that any integer. How do we know these numbers are multiples of 4? Also. 72 = 23 3. and this is best done with a simple example. For practice. then c is a multiple of d. 4. 14 = 2 7. is a multiple of 1 and n. 100 = 22 52. The GCF is 1. you could define a multiple as an integer with at least one factor. however. 3. 20. 12 = 22 3. and 96 are all multiples of 4. if their GCF is 1). All that really matters is that you understand the concept of multiples. Least Common Multiple The least common multiple (LCM) of two integers is the smallest multiple that the two numbers have in common. The GCF is 3. 49 = 72. Relatively Prime Numbers Two numbers are called relatively prime if they have no common prime factors (i. n. 4. what is the least common multiple of 4 and 6? We must first find their prime factorizations. 2. What are some multiples of 4? y 12. The GCF is 3 5 = 15. The LCM of two numbers is. for example. 45 = 32 5. There are no common prime factors. Page 27 . but neither number is prime. 27. 5. 80 = 24 5. like the GCF. 16 = 24. 12 and 32 15 and 26 34 and 40 3 and 17 18 and 16 Compare your answers to the solutions: 1. Page 28 Reducing Fractions . you get: As long as you multiply or divide both the numerator and denominator of a fraction by the same nonzero number. 40 = 23 5. you will not change the overall value of the fraction. It is the smallest prime factorization that includes 2 2 3. The LCM is 3 17 = 51. easier-to-work-with fractions. Reducing fractions makes life with fractions a lot simpler. 5. their LCM is 2 7 19 = 266. this is 2 2 3 = 12. To determine if two fractions are equivalent. The LCM is 24 32 = 144. A fraction describes a part of a whole. we start by finding the prime factorizations of both numbers: Therefore. Fractions Being able to efficiently and correctly manipulate fractions is essential to doing well on the Math IC test. 5. 12 = 23 3. It is composed of two expressions. 3. multiply the denominator and numerator of one fraction so that the denominators of the two fractions are equal. What is the LCM of 14 and 38? Again. Thus. 26 = 2 13. 4. a numerator and a denominator. The LCM is 23 5 17 = 680. The LCM is 25 3 = 96. The numerator of a fraction is the quantity above the fraction bar. For the numbers 4 and 6. Let¶s try a harder example. 2. Equivalent Fractions Two fractions are equivalent if they describe equal parts of the same whole. 3. For example. For some quick practice. and the denominator is the quantity below the fraction bar. 4. you will not change their fundamental relationship. 17 = 1 17. in the fraction 1 /2. 12 is the LCM of 4 and 6. Fractions represent a part of a whole. 18 = 2 32. For example. 2. 34 = 2 17. 1/2 = 3/6 because if you multiply the numerator and denominator of 1 /2 by 3. It takes unwieldy fractions such as 450 /600 and makes them into smaller. find the LCM of the following pairs of integers: 1. 3 = 1 3.Their LCM is the smallest prime factorization that contains every prime number in each of the two original prime factorizations. 15 = 3 5. so if you increase both the part and whole by the same multiple. 32 = 25. 1 is the numerator and 2 is the denominator. The LCM is 2 3 5 13 = 390. is to multiply each numerator by the same value as their respective denominator. We¶ll cross-multiply 200/20. comparing two fractions can be very simple. since 6 is the LCM of 2 and 3. For 1/2: So.000 and 2/3: Since 40. The first step is to make the denominators the same. However. The same process is repeated for the second fraction. then write the product of each multiplication next to the numerator you used to get it. such as 5. which is simply the LCM of the two denominators. For example. the new fraction is 3 /6.To reduce a fraction to its lowest terms. 2 /3: Page 29 . like 5.000 > 600. For example. for 450 /600.000 and 2/3. you¶ll most likely be dealing with two fractions that have different numerators and denominators. When faced with this situation. all you have to do is add up the numerators: Subtraction works similarly. Fractions with the Same Denominators Fractions can be extremely easy to add and subtract if they have the same denominator. the process becomes somewhat more involved. and other times you will have two fractions with different denominators. the GCF of 450 and 600 is 150. the fraction with the smaller denominator is bigger. A fraction is in reduced form if its numerator and denominator are relatively prime (their GCF is 1).000 might seem like a big. it makes sense that the equivalent fractions we studied in the previous section all reduce to the same fraction. divide the numerator and denominator by their GCF. Sometimes you will be given two fractions with the same denominator. then you simply subtract one numerator from the other: Fractions with Different Denominators If the fractions do not have equal denominators. Adding and Subtracting Fractions On SAT II Math IC. If the denominators of the fractions are equal. 200/ 2 20. If the denominators of two fractions are the same. the equivalent fractions 4/6 and 8/12 both reduce to 2/3. 2 /3 is the greater fraction. All you have to do is multiply the numerator of each fraction by the denominator of the other. For example.000. In certain cases. In addition problems. and then to subtract as described above. an easy way to compare these two fractions is to utilize cross-multiplication. the LCD of 1/2 and 2/3 is 6. such as 200/20. But fractions do not work the same way. If the numerators of the two fractions are the same.000.000. impressive fraction. 1/ 2 + 2 /3. are greater than numbers with fewer digits. Comparing Fractions When dealing with integers. The best way to do this is to find the least common denominator (LCD). The second step. but /3 is actually larger. after you¶ve equalized the denominators of the two fractions. Thus. then the fraction with the larger numerator is bigger. you will need to know how to add and subtract two different types of fractions. For example. Let¶s take a look at how to do this for our example. because 2 is a much bigger part of 3 than 200 is of 20. large positive numbers with a lot of digits. So the fraction reduces down to 3»4. 3/6 + 4/6 = 7/6. Here¶s a numerical example: Mixed Numbers A mixed number is an integer followed by a fraction. for a numerical example: Dividing Fractions Multiplication and division are inverse operations. The drawback to this second approach is that you will have to work with larger numbers and reduce your answer in the end. you simply divide the numerator of a fraction by the denominator. After all. For example. if the two denominators are 6 and 8. you could use 6 8 = 48 as a denominator instead of 24 (the LCD). subtraction. you multiply the integer portion of the mixed number by the denominator. . 2.The new fraction is 4 /6. and then multiply. The final step is to perform the addition or subtraction. making 3 the numerator of the improper fraction. Symbolically. that to perform division with fractions. It makes sense. Here¶s another example: Decimals Comparing Decimals Like fractions. First. all you have to do is flip the second fraction . For example. If you think it will be faster. Now. As a general rule. 1/2 = 1 2 = . which is also called taking its reciprocal. this can be represented as: Or. But. like 11/ 2. the product of the denominators will actually be the LCD (2 3 = 6 = LCD). multiplication. In some cases. It is another form of an improper fraction. when comparing two decimals such as . other times.3 with . But if asked Page 30 Decimals are just another way to express fractions. and you have your converted fraction. such as our example. The product of two fractions is the product of their numerators over the product of their denominators.003. or division can only be performed on the improper fraction form. and add that product to the numerator. you can always skip finding the LCD and multiply the denominators together to get a common denominator. In this case. the decimal with more leading zeroes is the smaller one. simply put 3 over the original denominator. But operations such as addition. So 1 2 + 1 = 3. to produce a decimal. which is a fraction greater than one. so you need to know how to convert between mixed numbers and improper fractions. Let¶s convert the mixed number 11 /2 into an improper fraction. the product of the denominators will be greater than the LCD. Multiplying Fractions Multiplying fractions is quite simple. then.5. comparing decimals can be a bit deceptive. and then you¶ll have to choose from fractions for test choices. Let the denominator be the number 1 followed by as many zeroes as there are decimal places in the decimal number. Whatever the case.25).0009. we put four zeroes in the denominator: Then. and because 9 is the larger integer. Use caution to avoid such mistakes. so when you attend school 25 percent of the time. we eliminate the decimal point and make 3875 the numerator: Since . by finding the GCF of 3875 and 10.000900 is smaller than . It might help to line up the decimal points of the two decimals: y .003 with . and then asks you to determine how much of that whole 3 represents in percentage form. y . it may just be easier to work with fractions. Reduce the fraction. Remove the decimal point and make the decimal number the numerator.0009 is clearly smaller than .3875 has four digits after the decimal point. To convert a decimal number to a fraction: 1. Percent literally means ³of 100´ in Latin. Sometimes you¶ll produce a decimal while solving a question.3875 into a fraction. 3. both conversions can be done easily. That would be wrong.´ to solve the question you have to set the fraction 3/ 15 equal to x»100: Page 31 . so don¶t get any ideas from our example. we can reduce the fraction: To convert from fractions back to decimals is a cinch. Other times.0030 Similarly. Let¶s convert . that means you only go to school 25/ 100 of the time (or . Simply carry out the necessary division on your calculator. which is 125. First. 2. you might be tempted to overlook the additional zero. such as for 3/5: Percents A percent is another way to describe a part of a whole (which means that percents are also another way to talk about fractions or decimals). 15.0009 as the larger decimal. You would probably fail all your classes if your attendance percentage was that low.000. Since a percent is ³of 100.to compare . choose . take a look at this question: 3 is what percent of 15? This question presents you with a whole.000925 Converting Decimals to Fractions Knowing how to convert decimals into fractions and fractions into decimals are useful skills. Instead. 4 has been cubed. in ab. Saying that a number is ³cubed´ means that it has been raised to the third power.e. Cube. a is multiplied by itself b times.You then cross-multiply and solve for x: Converting Percents into Fractions or Decimals You should be skilled at converting percents into fractions and decimals. i. that it has an exponent of 2. Exponents An exponent defines the number of times a number is to be multiplied by itself. where a is the base. The base refers to the 3 in 35. Percents relate to decimal numbers very simply and directly. and b the exponent. too. It is the number that is being multiplied by itself however many times specified by the exponent. i. In the expression 62.. For example: To convert from a decimal number to a percent. take the percentage number and place it as the numerator over the denominator 100.346% = . There are some other terms that you should be familiar with: y y y y Base. Saying that a number is ³squared´ means that it has been raised to the second power. Page Common Exponents 32 . To convert from a fraction back to a percent. In the expression 43. Exponent.. A percent is a decimal number with the decimal point moved two decimal places to the left.35. 58 percent is the same as 58/ 100. To convert from a percent to a fraction. 6 has been squared. that it has an exponent of 3. For example. Square.22346. move the decimal point two places to the right: On an even more simplistic level. for example. 235% = 2. In a numerical example. The exponent (or power) is the 5 in 35. we can just say that 50% = . The exponent tells how many times the base is to be multiplied by itself. the easiest method is to convert the fraction into a decimal first and then change the resultant decimal into a percent. because these problems will definitely come up on the Math IC test.5 or 22. 25 = 2 2 2 2 2. Percentages greater than 100 exist. An exponent can also be referred to as a power: a number with an exponent of 2 is raised to the second power.e. Multiplying and Dividing Numbers with Exponents To multiply exponential numbers or terms that have the same base. If you¶re dealing with algebraic expressions that have the same bases and exponents. you must expand the exponents to get (3 3 3) + (4 4). For example. then they can simply be added and subtracted. finally. Here is a list of squares from 1 through 10: Memorizing the first few cubes can be helpful as well: Finally. 3x4 + 5x4 = 8x4. to add 33 + 42. For example. add the exponents together: Page 33 . Knowing these regularly used exponents can save you the time it would take to calculate them during the test. you have to first find the value of each power.It may be worth your while to memorize a few common exponents before the test. such as 3x4 and 5x4. and then add the two numbers. 27 + 16 = 43. and then. the first few powers of two are useful for many applications: Adding and Subtracting Numbers with Exponents In order to add or subtract numbers with exponents. you get ± 8. you get a negative number. when you multiply the +4 by the next ±2. you multiply the ±8 by the last ±2 and get +16. Then. you get +4 because you are multiplying two negative numbers. In such cases. When you multiply the first two ±2s together. since you¶re once again multiplying two negative numbers. raise both the numerator and denominator to that exponent: Exponents and Negative Numbers As we said in the section on negative numbers. Raising an Exponent to an Exponent Occasionally you might encounter an exponent raised to another exponent. and when you multiply a negative number by a positive number. raise their product to that exponent: To divide exponential numbers raised to the same exponent. as seen in the following formats (32)4 and (x4)3. you get a positive number. just subtract the exponents. To multiply exponential numbers raised to the same exponent. raise their quotient to that exponent: If you need to multiply or divide two exponential numbers that do not have the same base or exponent. y Page 34 When you raise a negative number to an even-number exponent. To see why this is so. For example (±2)4 = 16. when you multiply a negative number by another negative number. you get a positive number. Finally. you¶ll just have to do your work the old-fashioned way: multiply the exponential numbers out and multiply or divide the result accordingly. let¶s break down the example. multiply the powers: Exponents and Fractions To raise a fraction to an exponent.To divide two same-base exponential numbers or terms. (±2)4 means ±2 ±2 ±2 ±2. since you are multiplying a positive number by a negative number. . These rules affect how negative numbers function in reference to exponents. which equals (± 2)3. Likewise. too. For example: Fractional Exponents Exponents can be fractions. we¶ll go over roots thoroughly in the next section. all you have to do is look at the example above and stop the process at ±8. which is the same as : . These rules can help a great deal as you go about eliminating answer choices and checking potentially correct answers. Special Exponents There are a few special properties of certain exponents that you also need to know. in this case 91»2. a slightly more complicated example: Page 35 . and you get a positive answer. it is called taking the root of that number or term. But the principle at work is simple. Don¶t worry if some of this doesn¶t quite make sense now. Zero Any base raised to the power of zero is equal to 1. To see why. For example. so we are just giving you a quick introduction to the topic now. Negative Exponents Seeing a negative number as a power may be a little strange the first time around. For a more familiar example.y When you raise a negative number to an odd power. Note. you should know that its value is 1. you could eliminate any answer choices that are positive. This expression can be converted into a more convenient form: Or. that 00 is undefinded. For example. For example: Or. you get a negative number. 21 = 2. One Any base raised to the power of one is equal to itself. If you see any exponent of the form x0. if you have a negative number raised to an odd power. on that same question. and anything under the radical. look at Fractional exponents will play a large role on SAT II Math IC. you know your answer is wrong. Any number or term raised to a negative power is equal to the reciprocal of that base raised to the opposite power. for example. however. 213 » 5 is equal to the fifth root of 2 to the thirteenth power: The symbol is also known as the radical. This can be helpful when you¶re attempting an operation on exponential terms with the same base. is called the radicand. When a number or term is raised to a fractional power. (±67)1 = ±67 and x1 = x. and so on. For example. . but there are also cube roots (numbers raised to 1»3). Here are a few examples: The same rules that apply to multiplying and dividing exponential terms with the same exponent apply to roots as well.With that. But it is often easier to work with roots in a different format. Here are some examples to firm up your knowledge: Roots and Radicals We just saw that roots express fractional exponents. fifth roots. fourth roots as . to square the number 3 is to multiply 3 by itself: 32 = 3 3 = 9. When a number or term is raised to a fractional power. For example. when squared. These roots of higher degrees operate the same way square roots do. Square roots are the most commonly used roots. cube roots are shown as . it follows that the cube root of 27 is 3. is equal to the given number. Roots are like exponents. In other words. only backward. and as the radicand. etc. the square root of a number is the number that. is 3. Each root is represented by a radical sign with the appropriate number next to it (a radical without any superscript denotes a square root). Because 33 = 27. fourth roots. you¶ve got the four rules of special exponents. the expression can be converted into one involving a root in the following way: with the sign as the radical sign. The root of 9. Look for yourself: Page 36 . this sort of simplification can make your calculations easier. only it¶s a little shorter. Look at the following expression: This is a pretty nasty product to find²even when you¶re using a calculator.1 E33. Not too shabby.1 1033.Just be sure that the roots are of the same degree (i. By approximating each number using scientific notation. we find that we were less than 1% off. A number written in scientific notation has two parts: 1. The power of 10 by which you must multiply the first number in order to get the larger number that is being represented. you are multiplying or dividing all square roots or all roots of the fifth power). Scientific Notation and Calculators On many calculators. Scientific Notation Scientific notation is a convention used to express large numbers.. A number between 1 and 10. The capital letter ³E´ has the same role as the ³ 10(power)´. In the following examples. we¶ll first write a number and then express it in scientific notation: Scientific notation is particularly useful when a large number contains many zeroes or needs to be approximated because of its unwieldy size. Logarithms Page 37 . scientific notation is written differently from what you¶ve seen here. Also.e. 2. note the way in which we combined the terms in the last example to make the multiplication a little simpler: In general terms: Often. Approximating quantities in scientific notation can prevent unnecessarily messy calculations. scientific notation allows you to work with numbers that might either be very tedious to manipulate or too large to fit on your calculator. In general. Instead of 3. we can make the problem a lot easier: When we compare this approximation to the actual product. your calculator might read 3. log2 8 = 3 because 23 = 8. Having defined logarithms in a sentence. it should be able to calculate logarithms with different bases. In the example above. This means that for the equation log4 16 = 2. and the exponents can be added. 2 is the base and 3 is the logarithm. once you get the hang of it. but less advanced calculators might not. logarithm. if you punched in LOG 16. you¶ll need to be able to manipulate logarithms within equations. a is the base. b is the exponent. the LOG button assumes a base of 10. as long as your calculator is scientific. and logarithms are often linked together. In fact. or log2 4. Keeping this in mind should help reduce the mystery that seems to surround logarithms. Some calculators can calculate a logarithm with any base you want. A logarithm is the power to which you must raise a given number. In this case. on the logarithm questions you¶ll see in the Algebra chapter. you¶ll realize that solving logarithmic equations is actually quite simple and easy. y Page 38 The Quotient Rule: when logarithms of the same base are divided. respectively. you won¶t be able to mentally calculate it²so the calculator becomes an important tool. the key thing to remember is that a logarithm problem is really an exponent problem. In general. you would get log10 16. . log4 16 = 2 because 42 = 16 and = 4. roots. So. For any of these types of questions. to equal another number. and exponent isolates these values. In particular. mostly in simple equation-solving problems (which we cover in the next chapter). let¶s show one symbolically. Each method provides a way to isolate one of the three variables in these types of equations. Finding the root. the base remains the same. The next three equations are equivalent: For example. you should know how to perform the basic operations on logarithms: y The Product Rule: when logarithms of the same base are multiplied. Calculate a few logarithms for practice: Operations on Logarithms You will rarely see a test question involving basic logarithms such as log10 100. The Math IC likes to use logarithms in algebra problems. called the base. You should now be able to see why the three topics of exponents. and x is the product. On your calculator. Logarithms and Calculators Unless the logarithm is a very simple one.Logarithms are closely related to exponents and roots. But there is one important thing you need to be aware of. For example. the exponents must be subtracted. If a is even and negative. 4. Page 39 .y The Power Rule: when a logarithm is raised to a power. 2. the exponent can be brought in front and multiplied by the logarithm. and c is even. Review Questions 1. This similarity results from the fact that logarithms are just another way to express an exponent. b is negative. Evaluate the expression (A) (B) (C) (D) (E) 4 6 8 for the value x = 2. You might have noticed how similar these rules are to those for exponents and roots. which of the following choices could be equal to a (A) (B) (C) (D) (D) b + c 2 + 1? ±71 ±16 0 4 9 What is the absolute value of the difference between the LCM and GCF of 24 and 42? (A) (B) (C) (D) (D) 18 162 174 498 1002 Which of the following fractions is not equivalent to the others? 3. (a b) + c2 must be positive and even. 3. C 3 = 6. because 25 = 32. Now you just need to find the absolute value 5. Since two positive even numbers will sum to a positive even number. too. 4. the end result is odd and positive. you found that they. you should have realized that they were both equivalent to 3/7. B Page Fractions are equivalent to each other if the numerator and denominator of one fraction can be multiplied by the same scalar. The square of an even number is even and positive. 2. exponents. If you checked the last two. B Find the prime factorizations of 24 and 42 to find their LCM and GCF.(A) (B) (C) (D) (D) 5. The numerator simplifies to The denominator is 5. and the product of an even number and any other number is even: therefore. are equal to 3/7. So the answer is 40»5 = 8. and the result is the other fraction. We¶ll solve it step by step. so c2 is even and positive. E The product of two negative numbers must be positive. After reducing the first two fractions. 9 is the only answer choice that is odd and positive. (a b) must be even and positive. D This question tests your understanding of order of operations. and logarithms. you could have stopped. The LCM of the two numbers is 23 3 7 = 168. 40 . and the GCF of the two numbers is 2 of the difference between the LCM and the GCF: |168 ± 6| = |162| = 162. As soon as you got to the third one and found that it was already in reduced form and not equal to 3 /7. When you add one to this value. (A) (B) (C) (D) (D) How many digits are in the number 5 3 3 ? 23 24 25 33 38 Explanations 1. it took him: to drive to Giambia City. the algebra tested on the math subject tests is not all that difficult. To find the duration of his flight. so there must be 23 + 1 = 24 digits in the full number. When you type 533 into your calculator. Jasonville. a tough problem that you are unsure how to solve might become easy if you try to plug in some answers. When you study your practice tests and look over the algebra questions you got wrong. Since traveling time = distance speed. Math IC Algebra Strategies There are several ways to answer most algebra problems. there is some good news. using algebra is probably the quickest method. This means that the Algebra THIS CHAPTER ON ALGEBRA IS A BEHEMOTH. It is by far the longest chapter in this book full of lengthy chapters. There¶s a reason for our extensive treatment: algebra is the most tested topic on the Math IC test. Before this information all starts to sound overwhelming. Let¶s use a sample algebra problem to illustrate these separate approaches: A baseball player travels from his home city. Only the topics you do need to know are covered in this chapter. the Math IC test-writers focus on a limited set of algebraic topics. None of these methods is necessarily better than the others. use scientific notation. For a problem you know how to solve.5 12 Using Algebra This question is a simple rate problem that can be solved with a few basic equations. If the distance from Jasonville to Giambia City is 250 miles.To answer this problem. After the game.5 4 4. it is approximately 1. Did you plug in answers when you should have used algebra? Did you use algebra when you should have plugged in answers? We¶ll lay out for you the different problem-solving approaches and tell you all you need to know about them. First. Then you can decide for yourself which method to choose. He drives at 50 miles an hour.16 decimal point has been moved over 23 decimal places. Second. you should think about the method you employed. to Giambia City for a baseball game. In contrast. 1023. you could try to avoid algebra and simply plug the answer choices back into the question until one of them works out. he travels back home and takes a flight that travels at 500 miles an hour. Alternatively. and some of the questions that focus on geometry or trigonometry still involve some sort of algebraic technique or concept. About 30 percent of the Math IC questions directly test your algebraic abilities. Remain flexible in your approach to each question and choose the method that best suits the problem. we use the same rate formula: Page It took the player: 41 . and it took him j hours longer to drive than to fly. Or you can pick numbers to substitute into the various expressions given as answer choices. what is j? (A) (B) (C) (D) (D) 1 3. You could try to solve a problem by using standard algebra and setting up and solving an equation. Picking Numbers Picking numbers is a variation of plugging in and should only be used when the answer choices contain variables. these numbers aren¶t realistic (who flies at 10 miles an hour?). v = 100. it takes him 20 ± 10 = 10 hours longer to drive. Picking numbers allows you to transform variables into concrete numbers. The same answer choice will always surface as long as you plug in consistently and follow all guidelines given by the problem. we try D 4. First. A modified version of our original sample question shows what kind of problems might lend themselves to picking numbers. and see whether the given information holds true. To use the picking numbers method. Clearly. Using our numbers. in the baseball player problem.5. A baseball player travels from his home city. but your goal is to pick easy-to-manipulate numbers. let m = 5. After the game. In such instances. if it takes him C 4 hours more to drive. and p into all the answer choices. Jasonville.5 = . v. he travels back home. to solve the question: it takes the baseball player 250 50 = 5 hours to drive to Giambia City. plugging in might be the best method for you. you should make full use of the fact that the answer choices on Math IC are always presented in ascending value. than it takes him 5 ± 4 = 1 hour to fly back to Jasonville.longer to drive. After plugging m. But thinking in terms of variables can be confusing to some people. He drives at m miles an hour. It doesn¶t matter what specific numbers you plug into a problem. especially as you come across the more difficult questions at the end of the test. Plugging In Answers Sometimes you might not be sure how to approach a problem or don¶t have the time to think out the proper equations. and takes a flight instead at p miles an hour. it takes the baseball player 100 5 = 20 hours to drive and 100 10 = 10 hours to fly. Therefore. 42 . So.5 = 250 miles on his flight. we find that only D produces an answer of 10. For example. So. since if it doesn¶t turn out to be the answer. You¶re essentially testing the relationships between the variables in each given answer and ensuring they remain true. D is the answer. and it took him j longer to drive than to fly. If the distance from Jasonville to Giambia City is v miles. It takes him 5 ± 4. what is j? (A) (B) (C) (D) (D) Page This question asks you to figure out which set of variables in the answer choices is the right one. it must take him longer than 4 hours more to drive than to fly.5 hours to fly. Now. All you have to do is substitute the answer choices back into the problem. Next. D is the correct answer. to Giambia City for a baseball game. you can usually tell whether to try a smaller or larger answer choice. and p = 10. he could fly 500 miles. But the question tells us that in 1 hour. The process of plugging in is simple. you need to select numbers and plug them into the answer choices. which means that he travels 500 . So start by plugging in answer choice C. A lot of the algebra that you¶ll have to perform on the SAT II Math tests will consist of solving an equation with one variable. like apples or dollars. Most of this chapter. Equation. Other times. First. this will slow you down.Very rarely. Page For some questions on the Math IC test. avoid 0. For constant 4. ±7x. but it is time-consuming. deals with different techniques for simplifying expressions and solving different types of equations. Variables will sometimes represent specified quantities. we¶ll review the algebra topics covered in the Math IC Subject Test. which is what differentiates expressions from equations. Equation-Solving There are a number of algebraic terms you should know in order to be able to talk and think about algebra: y y y Variable. in fact. Some methods work best some times. such as {(x2 + 2)3 ± 6x} » 7x5. are considered terms because they are considered coefficients of variables raised to the zero power. Or an expression can be as complicated as the sum or difference of many terms. when you are picking numbers. in the equation below. The letters x and y are the most commonly used letters for variables. or any numbers that appear in the answer choices. more than one answer choice will result in the correct answer for the first set of numbers you picked. but a variable can be represented by any letter in the English alphabet. You will almost never have to plug in more than two sets of numbers. a number. Another way to define a term is as any quantity that is separated from other quantities by addition or subtraction. the left side contains four terms {x3. Now. The constants. including constants. like 5. we¶ll review how to write an equation. Part of your practice for the Math IC test will be to get comfortable with algebra questions so that you can choose which method you want to use for every question. An unknown quantity. The product of a constant and a variable. 4 = 4x0. The Bottom Line As you can see. but that¶s the price you pay for using this method. for example. We¶re talking about the language 43 Writing Equations . 4 and ±1. 4} and the right side contains two terms {x. Finally. y y Expression. and it also allows you to check your math for careless calculations. 2x2. A quantity that does not change. An expression can be as simple as a single constant term. they can only be simplified. you¶ll need to translate the problem from a language you¶re used to²English²into a more useful. When this occurs. written as a letter. Constant. You¶ll need to manipulate variables just to show that you understand certain algebraic principles. In other words. Term. albeit less familiar language. So every term. a specific meaning won¶t be attached to them. there is no ³right´ method to solving all algebra problems. Obviously. Expressions therefore cannot be solved. For example. Expressions don¶t include an equal sign. Greek letters are also used quite often. Any combination of terms. 1. and others work best at other times. each of which is a combination of constants and variables. When picking numbers. Two expressions linked by an equal sign. Picking these numbers can overly simplify the expressions you are dealing with and cause you to pick the wrong answer. Picking numbers gives you a mechanical method of solving tricky problems. is the product of a constant and a variable raised to some power. you must check through all the answer solutions with your chosen numbers. simply plug in a different set of numbers. ±1}. They figure that a good number of test-takers will see only the word price. All of the marbles in the sack are either red or blue. not an equation) for the price in dollars of 35 oranges. of course. find the price of those oranges: But wait. According to the problem. Next. note that the variable r didn¶t appear anywhere in the answer. Did you notice that the question asked for the price of 35 oranges in dollars? The writers of the Math IC are a clever bunch. Don¶t worry about the solution for now²just focus on how we translated the word problem into equations that lead to the solution. The best way to learn how to do these things quickly and effectively is to practice. and they will not notice what units are asked for. if 35 >r.of math. we need to find an expression (notice. We know there are 100 cents per dollar. where r is the number of red marbles. but the first f oranges are free (f < r). Find an expression for the price in dollars of 35 oranges. which we¶ll cover in the coming sections. Be careful not to fall into their carefully laid trap. For example. You could also write b = r ± 20 to signify the same concept. if not a little sneaky. That problem was easy. Let¶s list the two equations we have so far: Using both of these equations. This problem gives a clear-cut request: how many blue marbles are in the sack? You must therefore find the value of b. you can write r + b = 50. After a little manipulation. This part of the word problem can be written in the form of an equation as r = b + 20. you need to decipher what exactly the question is asking for. you¶ll simply be asked to find an expression for a certain quantity described in a word problem. More information needs to be incorporated. you can solve for b. you can¶t do that with just this equation. Egad! It is yet another attempt (and a common one at that) by those devious test-writers to lower 44 . In other cases. The key to a problem like this one is working step by step. This equation tell us that all of the 50 marbles in the sack are either red or blue. Unfortunately. Here¶s a sample problem: In a sack of 50 marbles. Here¶s a harder one: Stan sells oranges for c cents apiece. there are 20 more red marbles than blue marbles. The minimum number of oranges that Stan will sell to an individual is r. and 35 >f. First. and b is the number of blue marbles in the sack. Now that we have a starting equation. How many blue marbles are in the sack? To start with. you¶ll find that b = 15 (and r = 35). so we can easily convert the price by dividing by 100: Page Before we move to another problem. find out how many of the 35 oranges aren¶t free of charge: because f is the number of oranges that are free. and one of your major test-taking responsibilities is being able to write an equation based on the pertinent information you¶re given by a problem. use the knowledge that there are 20 more red marbles than blue marbles. Thus. But the equation will always remain true as long as you always do the same thing to both sides. but you need to carry out just four steps to solve it: 1. take the square root of the other. Gus also needs to buy ten pairs of new jeans (he is uncoordinated and spills often). Subtract 2 from both sides: 1. what is the difference (in dollars) between the cost of the paint and the cost of the jeans? Assume he doesn¶t buy any excess paint²that is. You will. especially word problems. don¶t worry. Here¶s one last problem: Gus needs to paint his house. You may come across many problems. 4. in dollars. between the cost of the paint and the cost of the jeans is xp/ y ± 10d. let¶s look at what happens when you manipulate the equation 3x + 2 = 5. The brand of paint he buys (at a cost of p dollars a can) comes in cans that cover y square feet each. which has a surface area of x square feet. of course. The jeans Gus buys cost 10d dollars. 1. If you take the square root of one side of an equation. If you divide one side of an equation by 3. If you¶re still uncomfortable doing this. in which extraneous information is provided only to confuse you. 3. change the form of the equation²that¶s the point of manipulating it. Multiply both sides by 2: Page 45 . the difference. the required amount is not a fraction of a can. Gus must buy x/ y cans of paint to cover his house. with x = 1. Just because a variable or number appears in a problem doesn¶t mean that it will be useful in finding the answer. This will cost him xp/ y dollars. 2. the next thing to do is to solve for the value that the question asks for. For the rest of this chapter. First and foremost. we¶ll constantly be converting word problems into equations.your score. For example. you must divide the other side by 3. This word problem is long and complicated. the most important thing to remember when manipulating equations is to do exactly the same thing to each side of the equation. You¶ll get a lot more practice in the sections to come Manipulating Equations Now that you know how to set up the equation. They cost d dollars a pair. If Gus makes these purchases. By treating the two sides of the equation in the same way. you can rest easy that you won¶t change the meaning of the equation. The idea is to ³undo´ everything that is being done to the variable so that it will be isolated in the end. And finally.´ For the quickest results. the variable x is being squared. take the equation apart in the reverse order of operations. Add 4 to both sides: These examples show that you can tamper with the equation in any way you want. you can manipulate the question how you want without affecting the value of its variables. first add and subtract any extra terms on the same side as the variable. We need to do the opposite of all these operations in order to isolate x and thus solve the equation. and you have successfully ³solved for the variable. multiply both sides of the equation by 4: Next. added to 5. Let¶s look at an example: In this equation. by definition. If you follow this rule. that variable is equal to everything on the other side. That is. Solving an Equation with One Variable To solve an equation with one variable. This process is PEMDAS in reverse (SADMEP!). as long as you commit the same tampering on both sides. First. Next. subtract 1 from both sides of the equation: Then. subtract 5 from both sides of the equation: Again. multiply and divide anything on the same side of the variable. you must isolate that variable. etc. divide both sides of the equation by 3: Now. raise both sides of the equation to a power or take their roots according to any exponent attached to the variable. do anything inside parentheses. Then. divide both sides of the equation by 3: Page 46 . Then.1. Isolating a variable means manipulating the equation until the variable is the only thing remaining on one side of the equation. multiplied by 3. performing an operation on a variable is mathematically no different than performing that operation on a constant or any other quantity. In the next section. The key step is to multiply both sides by x to extract the variable from the denominator. Here¶s another. Equations like these are solved the same way as any other equation. two techniques that were used in this example. slightly more complicated example: This question is a good example of how it¶s not always simple to isolate a variable. (Don¶t worry about the logarithm in this problem²we¶ll review these later on in the chapter. It is not at all uncommon to have to move the variable from side to side in order to isolate it.Finally. as you can see. except that you may need different techniques to isolate the variable.) However. even the thorniest problems can be solved systematically²as long as you have the right tools. it might be in a denominator or an exponent. we¶ll discuss factoring and distributing. Let¶s look at a couple of examples: Solve for x in the equation + 2 = 4. Sometimes the variable that needs to be isolated is not conveniently located. For example. take the square root of each side of the equation: We have isolated x to show that x = ±5. So. we¶ll reemphasize two things: Page 47 . Remember. having just given you a very basic introduction to solving equations. In the last example from the previous section on manipulating equations. But if you distribute the 3y. The basis for both techniques is the following property.1. from a variable to a constant to a combination of the two. consider the expression 3y(y2 ± 6): If we set the original. you can see why distributing facilitates the solving of some equations. So it follows that you can factor or distribute one side of the equation without doing the same for the other side of the equation. Now we get into some more interesting tools you will need to solve certain equations. called the distributive property: Similarly: a can be any kind of term. Distributing When you distribute a factor into an expression within parentheses. you simply multiply each term inside the parentheses by the factor outside the parentheses. undistributed expression equal to another expression. Do the same thing to both sides. For example. Consider the expression 4x3 ± 8x2 + 4x. we distributed and factored to solve an equation. Work backward (with respect to the order of operations). for example. which is 4x: The expression simplifies further: Page See how useful these techniques are? You can group or ungroup quantities in an equation to make your calculations easier. We later factored the term x out of the expression x log 2 ± x log 3 (on the left side of the equation). First. you get: Subtracting 3y3 from both sides gives us: Factoring Factoring an expression is essentially the opposite of distributing. 48 . You can factor out the GCF of the terms. 2. Distributing and Factoring Distributing and factoring are two of the most important techniques in algebra. Solving 3y (y2 ± 6) = 3y3 + 36 looks quite difficult. They give you ways of manipulating expressions without changing the expression¶s value. we distributed the quantity log 3 into the sum of x and 2 (on the right side of the equation). Generally speaking. and factoring creates them. you know that either: 1. Let¶s see a few examples: Combining Like Terms After factoring and distributing. It¶s your job as a Math IC mathematician to decide which technique will best help you solve a problem. you know that at least one of the terms is equal to zero. For example. you can factor out the variable and add or subtract the coefficients. A general formula for combining like pairs looks something like this: Zero Product When the product of any number of terms is zero. x = y = 0. Consider this equation: Again. if xy = 0. x = 0. combining them into one coefficient and therefore combining the ³like´ terms into one term. and involves adding or subtracting the coefficients of variables that are raised to the same power. y = 0.Distributing eliminates parentheses. since 3x2 or (x + 2) must equal 0. since one of the expressions in parentheses must be equal to 0. the expression: can be simplified to: by adding the coefficients of the variable x3 together and the coefficients of x2 together. when you have an expression in which one variable is raised to the same power in different terms. Page 49 . either x = ±4 or x = 3. by combining like terms. we know that either x = 0 or x = ±2. there are additional steps you can take to simplify expressions or equations. and x  0 3. and y  0 2. This is useful in a situation like the following: In this equation. Combining like terms is one of the simpler techniques you can use. For example. Inequalities Page 1. 2}. Absolute Value To solve an equation in which the variable is within absolute value brackets. In this case. First. x > y: ³x is greater than y. we should introduce inequalities. So the solutions to the equation |x + 3| = 5 are x = {±8. Keep one of these two equations the same. x = ±8. while in the other negate one side of the equation. This is why there are always two solutions to absolute value problems (unless the variable is equal to 0). first isolate the expression within the absolute value brackets and then divide the equation into two. Here is one more example: Solve for x in terms of y in the equation 3 = y 2 ± 1. solve for x in the equation x + 3 = 5. solve for the variable as if the expression within absolute value brackets were negative: The solution set for x is {y2 ± 3. you must divide the equation into two equations.Keep your eye out for a zero product²it¶s a big time-saver. An inequality is like an equation. In this case. either x = c or x = ±c. There are four types of inequalities: . the absolute value of the expression within brackets will be the same. x = 2.´ 50 Before you get too comfortable with expressions and equations. especially when you have multiple-choice answers to choose from. Generally speaking. In this case. but instead of relating equal quantities. it specifies exactly how two quantities are not equal. A slightly more complicated example is this: In this problem. Then. In either case. to solve an equation in which the variable is within absolute value brackets. solve for x in the equation x + 3 = ±5. ±y2 ±1}.´ 2. x < y: ³x is less than y. isolate the expression within the absolute value brackets: Then solve for the variable as if the expression within absolute value brackets were positive: Next. The most basic example of this is an equation of the form |x| = c. you must solve two equations: First. this idea makes sense. Solve for x in the inequality |2x ± 4| ” 6. Solve for x in the inequality • ±2. remember that if x > y. with a lower and an upper bound. y First. x • y: ³x is greater than or equal to y. then the solution is a single range. Notice that in the last example. the inequality had to be reversed. 1. the solutions come in two varieties. then ±x< ±y. For example. x ” y: ³x is less than or equal to y. just as 5 > 4 and ±5 < ±4. Intuitively.´ 4. Absolute Value and Inequalities When absolute values are included in inequalities. solve for the upper bound: y Second. solve for the lower bound: Page 51 . and it might help you remember this special rule of inequalities. Here are a few examples: Solve for x in the inequality ± 3 < 2y. If the absolute value is less than a given quantity. the direction of the inequality switches.3. To help remember that multiplication or division by a negative number reverses the direction of the inequality. Another way to express the solution is x • ±2.´ Solving inequalities is exactly like solving equations except for one very important difference: when both sides of an inequality are multiplied or divided by a negative number. 1. Solve for x in the inequality |3x + 4| > 16. The range of a does not include its lower bound (it is appropriate for people ³older than 40´). solve for the lower range: y Now combine the two ranges to form the solution. This occurs when the absolute value is greater than a given quantity. i.. Ranges Inequalities are also used to express the range of values that a variable can take.y Now. For example.e. which is two disjoint ranges: ±’ <x< ±20»3 or 4 <x< ’. it is important to first isolate the expression within absolute value brackets. and one whose lower bound is a real number and whose upper bound is infinity. The lower bound of a is 40. The other solution for an absolute value inequality involves two disjoint ranges: one whose lower bound is negative infinity and whose upper bound is a real number. and only then. should you solve separately for the cases in which the quantity is positive and negative. Then. Consider the following word-problem example: A very complicated board game has the following recommendation on the box: ³This game is only appropriate for people older than 40 but no older than 65. When working with absolute values. a < x < b means that the value of x is greater than a and less than b. 65 is appropriate. solve for the upper range: y Then. y First. but it does include its upper bound (³no older than 65´. combine the two bounds into a range of values for x. and the upper bound is 65. ±1 ” x ” 5 is the solution. but 52 .´ What is the range of the age of people for which the board game is appropriate? Page Let a be the age of people for which the board game is appropriate. what is the range of x + y? Page 53 There is one crucial rule you need to know about multiplying ranges: if you multiply a range by a negative number. you must flip the greater-than or less-than signs. small errors occur on virtually every part.5 grams. The answer to the problem is 21. ADDITION WITH RANGES OF TWO OR MORE VARIABLES .98 21.5 = 21.93 grams.5 = 21. If the target weight of this piece is 21. the range of the age of people for which the board game is appropriate can be expressed by the inequality: Here is another example: A company manufactures car parts. The problem states that the piece cannot weigh less than the minimum weight or more than the maximum weight in order for it to work. simply manipulate the range like an inequality until you have a solution. and multiplying two-variable ranges. what is the range of 2x + 3? To solve this problem. The company knows that a particular piece they manufacture will not work if it weighs less than 98% of its target weight or more than 102% of its target weight. if you multiply the range 2 <x < 8 by ±1. For example: If 4 <x < 7. subtraction. Therefore. In general. Math IC questions that ask you to perform operations on ranges of one variable will often test your alertness by making you multiply the range by a negative number. This means that the part will function at boundary weights themselves. Every time you come across a question involving ranges. you should carefully peruse the problem to pick out whether a particular variable¶s range includes its bounds or not.´ Operations on Ranges Operations like addition.02 21. Finding the range of a particular variable is essentially an exercise in close reading. As is the case with any system of mass production. and the lower and upper bounds are included. Begin with the original range: Then multiply the inequality by 2: Add 3 to the inequality. The key for this company to succeed in making viable car parts is to keep the errors within a specific range.07 ” x ” 21. and you have the answer: If ±2 <x< 8 and 0 <y< 5. This inclusion is the difference between ³less than or equal to´ and simply ³less than. in what range of weights must the piece measure for it to function? The boundary weights of this car part are . Some range problems on the Math IC will be made slightly more difficult by the inclusion of more than one variable. the same basic procedures for dealing with one-variable ranges applies to adding.66 is not). For instance. where x is the weight of the part in grams. the new range will be ±2 >x> ±8. subtracting. and multiplication can be performed on ranges just like they can be performed on variables.07 and 1.93. to subtract ranges. ±2 <x + y< 13. is the range 0 <x + y< 11.Simply add the ranges. and then add the ranges as usual. 4 + 1 = 5. what is the range of jk? First. but not its upper bound. The range of x includes its lower bound. or some combination of these two? The rule to answer this question is the following: if either of the bounds that are being added. and the range of 2(x + y) is 0 ” 2(x + y) < 22. or multiplied are inclusive (” or •) is the resulting bound also inclusive. and the greatest is the upper bound. 7. what is the range of 2(x + y)? The first step is to find the range of x + y. you have to find the range of ±t. Rewrite it first: Next add the ranges to find the range of x + y: Page Systems of Equations 54 We have our bounds for the range of x + y. The upper bound is 8 + 5 = 13. multiply the lower bound of one variable by the lower and upper bounds of the other variable: Then. The lower bound is ±2 + 0 = ±2. ±12 <jk< 48. What is the range of s ± t? In this case. Only when both bounds being added. By multiplying the range of t by ±1 and reversing the direction of the inequalities. . Therefore. the range of x + y is 0 ” x + y< 11. 5 <s ± t< 10. or multiplied is non-inclusive (< or >). subtracted. 3. then the resulting bound is non-inclusive. In general. Therefore. multiply the upper bound of one variable with both bounds of the other variable: The least of these four products becomes the lower bound. and 7 + 3 = 10. Notice that the range of y is written backward. Let¶s try one more example of performing operations on ranges: If 3 ” x< 7 and . but are they included in the range? In other words. 0 ” x + y ” 11. subtracted. SUBTRACTION WITH RANGES OF TWO OR MORE VARIABLES Suppose 4 <s < 7 and ±3 <t < ±1. The range of y includes both its bounds. MULTIPLICATION WITH RANGES OF TWO OR MORE VARIABLES If ±1 <j< 4 and 6 <k< 12. we find that 1 < ±t< 3. find the range of the opposite of the variable being subtracted. Therefore. Now we can simply add the ranges again to find the range of s ± t. Therefore. with the upper bound to the left of the variable. Solve for x in terms of k. Instead. Here is a slightly more complicated example. and then x = y ± 3 + 4 = 3 ± 3 + 4 = 4. easier type involves substitution. and the second involves manipulating equations simultaneously. there are essentially two types of systems of equations that you will need to be able to solve. you cannot solve for x in terms of k using just the first equation. To solve for x. and it must come in the form of another equation. The first. Say. for example. Suppose 3x = y + 5 and 2y ± 2= 12k. then y = 3. The second contains only y. you must solve for y in terms of k in the second equation. But given another equation with the same two variables x and y. you can find the value of x by adding the two equations together: Page Here is another example: 55 . For the Math IC. you must solve for y in the second equation and substitute that value for y in the first equation. Simultaneous Equations Simultaneous equations refer to equations that can be added or subtracted from each other in order to find a solution. These multiple equations containing the same variables are called systems of equations. Substitution Simply put. we have two equations. If 2y = 6. Consider the following example: Suppose 2x + 3y = 5 and ±1x ± 3y = ±7. The first equation contains x and y. Additional information is needed. that a single equation uses the two variables x and y. what is x? In this case.Sometimes. and then substitute that value in the first equation to solve for x. substitution is when the value of one variable is found and then substituted into the other equation to solve for the other variable. Again. and using the methods we¶ve discussed up until now will not be enough to solve for the variables. Then substitute y = 6k + 1 into the equation 3x = y + 5. Try as you might. It can be as easy as this example: If x ± 4 = y ± 3 and 2y = 6. you won¶t be able to solve for x or y. then the values of both variables can be found. What is x? In this particular problem. a question will have a lone equation containing two variables. and manipulate the equations to find the solution. the answer is given. Simultaneous equations on the Math IC will all be this simple. Once you know the independent values of x and y. Then. 4x is simply twice 2x. as above.6x + 2y = 11 and 5x + y = 10. for example. What is x + y? By subtracting the second equation from the first: Some test-takers might have seen this problem and been tempted to immediately start trying to solve for x and y individually. Common Word Problems The writers of the Math IC love word problems. so by multiplying the first equation by 2. 2 (2x + 3y = ±6) = 4x + 6y = ±12 Now add the equations and solve for y. the key to a rate problem is correctly placing the given information in the three categories. and we have the nitty-gritty on all of them. you can multiply them together. work. and price. For example. you could solve for y. the Math IC uses only a few types of word problems. Luckily. like we have in this last example. you can substitute the values into the rate formula. These problems force you to show your range as a mathematician. They demand that you read and comprehend the problem. speed is a rate that relates the two quantities of distance and time. Only as a last resort should you solve for one variable in terms of the other and then plug that value into the other equation to solve for the second variable. which means that you should find a way to eliminate one of the variables by adding or subtracting the two equations. They will have solutions that can be found easily by adding or subtracting the equations given. you can then add the equations together to find y. Rates A rate is a ratio of related qualities that have different units. you can solve for the second variable using either of the original equations. If the last question had asked you to calculate the value of xy. and then solve for x by substitution into either equation. We¶ll look at the three most common types of rate: speed. Give this last example a try: 2x + 3y = ±6 and ±4x + 16y = 13. When you solve for one variable. set up an equation or two. . What is the value of y? The question asks you to solve for y. The better test-taker notices that by subtracting the second equation from the first. Here is the general rate formula: Page Speed 56 No matter the specifics. One morning. how long will it take him to finish the race? Immediately. the information that you¶ll need to solve the problem will often be given in a less straightforward manner. time is quantity a and distance is quantity b. How many hours did he rollerblade? This question provides more information than simply the speed and one of the quantities.280 feet in a mile. each representing a state. Jim starts rollerblading and doesn¶t stop until he has gone 60 miles. since the hour in the rate is at the bottom of the fraction. Here¶s an example: Jim rollerblades 6 miles per hour. Sometimes a question will give you inconsistent units. but basically we plugged some numbers into the rate equation and got our answer. Because rate questions are in the form of word problems.000 feet per hour. But you can be sure that the Math IC test won¶t simply give you one of the quantities and the rate and ask you to plug it into the rate formula. Always read over the problem carefully and don¶t forget to adjust the units²the answer choices are bound to include nonadjusted options.In the case of speed. then: Note that the hour units canceled out. If the race is 480 miles long. For example. just to throw you off. you should pick out the given rate of 528.000 feet per hour and notice that 480 miles are traveled. You should also notice that the question presents a units problem: the given rate is in feet cycled per hour. y y y Time a: x hours rollerblading Rate: 6 miles per hour Quantity b: 60 miles So. This problem requires a little analysis. The cyclist from California can cumulatively cycle 528. if you traveled for 4 hours at 25 miles per hour. We know unnecessary facts such as how Jim is traveling (by rollerblades) and when he started (in the morning). Here¶s a slightly more difficult rate problem: At a cycling race. we can find the rate for miles per hour: We can now plug the information into the rate formula: y y y Page 57 Time: x hours cycling Rate: 100 miles per hour Distance: 480 miles . since we know there are 5. like in this example. we can write: Jim was rollerblading for 10 hours. there are 50 cyclists in all. For this question. Ignore them and focus on the facts you need to solve the problem. and the distance traveled is in miles. Take a look at this sample problem: A professional golfer usually has an average score of 72. What is his new average? Page 58 . and you knew that each basketball cost $25 each: Percent Change In percent-change questions. if you knitted for 8 hours and produced two sweaters per hour. and the rate in price per item. Work In work questions. you will need to determine how a percent increase or decrease affects the values given in the question. Just remember the classic rate formula and use it wisely. you will be given one of the values and be asked to find the percent change. you will usually find the first quantity measured in numbers of items. and you will have to find either the original value or new value. Sometimes you will be given the percent change. but he recently went through a major slump. How long would it take for 8 men to dig a 60 foot well? Assume that these 8 men work at the same pace as the 4 men. The work done by the 8 men is 60 feet. The rate of 10 feet per day per 4 men converts to 20 feet per day per 8 men. and they work at a rate of 10 feet per day per 4 men. For example. We need to create our own rate. Can we use this information to answer the question? Yes. which is the size of the new crew. Price In rate questions dealing with price. the second quantity measured in work done. We are given a quantity of work of 40 feet and a time of 4 days. using whichever units might be most convenient. The group of 4 men dig 40 feet in 3 days. His new average is 20 percent worse (higher) than it used to be. you find that the group of 4 digs at a pace of 10 feet per day. we know that 8 men dig a 60 foot well. Now we use the rate formula: y y y Time:x days of work Rate: 20 feet per day per eight men Total Quantity: 60 feet This last problem required a little bit of creativity²but nothing you can¶t handle. From the question. let¶s examine what that problem says: 4 men can dig a 40 foot well in 4 days.8 hours to finish the race. you will usually find the first quantity measured in time. First. Other times. then: Here is a sample work problem. to carry over to the 8-men problem.So it takes the cyclist 4. Let¶s say you had 8 basketballs. It is one of the harder rate questions you might come across on the Math IC: Four men can dig a 40 foot well in 4 days. Dividing 40 feet by 4 days. the second measured in price. and the rate measured in work done per time. 4. Double Percent Change A slightly trickier version of the percent-change question asks you to determine the cumulative effect of two percent changes in the same problem. What is the final sale price of the bike? One might be tempted to say that the bike¶s price is discounted 30% + 20% = 50% from its original price. you know that his new score is 120% higher than his old score. Then. you have the original price and the sale price and need to determine the percent decrease. the price of the bike drops 30 percent: The second reduction in price knocks off an additional 20 percent of the sale price. its price is reduced by an additional 20%. higher price. In this case. By what percentage did its price drop? In this case. and then lowered 20%. Therefore.This is a percent-change question in which you need to find how the original value is affected by a percent increase. you can skip a step and multiply 72 1.3. Why? Because after the price was increased by 20 percent. the shirt¶s price was reduced by 20 ± 14 = 6 dollars. Here¶s another example of a percent-change problem: A shirt whose original price was 20 dollars has now been put on sale for 14 dollars. lower price²not the original amount. What is the final selling price of the computer? Page If this question sounds too simple to be true. For example.2 = 86. the reduction in price was a reduction of 20 percent of a new. Let¶s work through the problem carefully and see. After the first sale. So. For example: A bike has an original price of 300 dollars. then you should add it to his original average. the second percent decrease is 20 percent of a new.2. Its price is raised 20%. a 30% drop in the price of the shirt. If you see this immediately. Watch and learn: 59 . two weeks later. but the key to solving double percent-change questions is to realize that each percentage change is dependent on the last. Its price is reduced by 30%. it probably is. in the problem we just looked at. The final price is not the same as the original. Since you know that the golfer¶s score went up by twenty percent over his original score. to answer this question. 6 20 = . the final price will be lower than the original. not the original price: The trickiest of the tricky percentage problems go a little something like this: A computer has a price of 1400 dollars.20 to see what the change in score was: Once you know the score change. All you need to do is divide the amount by which the quantity changed by the original quantity. First. you should multiply 72 by . since his new average is higher than it used to be: It is also possible to solve this problem by multiplying the golfer¶s original score by 1. quantities like the one described in this problem are said to be growing exponentially. you need to perform percent-change operations repeatedly. you might start by calculating the population after one year: Or use the faster method we discussed in percent increase: After the second year. how great will the population be in 50 years? To answer this question. In questions involving populations growing in size or the diminishing price of a car over time. 50 separate calculations. Exponential decay is a repeated percent decrease. Solving these problems would be time-consuming without exponents. But solve it step by step. after the price is reduced by 20%: Double percent problems can be more complicated than they appear. the population will have grown to: And so on and so on for 48 more years. and you¶ll do fine. Let¶s work through a few example problems to get a feel for both exponential growth and decay problems. as well as the decay of physical mass. But instead of a quantity growing at a constant percentage. To calculate exponential decay: Page The only difference between the two equations is that the base of the exponent is less than 1. Here¶s an example: If a population of 100 grows by 5% per year. You may already see the shortcut you can use to avoid having to do. the quantity shrinks at a constant percentage. in this case. Exponential Growth and Decay These types of word problems take the concept of percent change even further. That is why the formulas that model these two situations are so similar. The formula for calculating how much an exponential quantity will grow in a specific number of years is: Exponential decay is mathematically equivalent to negative exponential growth. Exponential decay is often used to model population decreases. The final answer is simply: In general.Now. 60 . because during each unit of time the original amount is reduced by a fixed percentage. Chris¶s account will have $2500 1. To solve the problem.0515 § $4157. The rate is . and the time is 10. how many are there after 6 hours? The question. compounded monthly. Thus.0415 § $4502. and the time is 6 hours: Simple Exponential Decay Problem A fully inflated beach ball loses 6% of its air every day. our answer is: Here¶s another compounding problem: Sam puts 2000 dollars into a savings account that pays 5% interest compounded annually. It grows at a constant rate of 2% per year.36 in it.06. Plugging the information into the formula: More Complicated Exponential Growth Problem A bank offers a 4. with its growing population of bacteria. however. how much money will the account hold two years later? This problem is a bit tricky for the simple reason that the interest on the account is compounded monthly. If the beach ball originally contains 4000 cubic centimeters of air. there will be 2 12 = 24 compoundings of interest.Simple Exponential Growth Problems A population of bacteria grows by 35% every hour.7% interest rate on all savings accounts. Consider the following example: The population of a small town is 1000 on January 1. If the population begins with 100 specimens. Chris puts 2500 dollars into a different savings account that pays 4% annually. makes it quite clear that this is an exponential growth problem. This means that in the 2 years that question refers to. The decay rate is . that Sam¶s account is gaining on Chris¶s account. So. the original amount is 100. whose account will have more money in it. In what year does the population of the town first exceed 1500? Page 61 . 2001. Logarithms Logarithms have important uses in solving problems with complicated exponential equations. The time variable in the equation is affected by these monthly compoundings: it will be 24 instead of 2.85 in it after 15 years.035. the original amount is 4000 cubic centimeters of air. we know this is an exponential decay problem. After 15 years. Chris¶s account will still have more money in it after 15 years. Notice. you just need to plug the appropriate values into the formula for a repeated percent increase. how many cubic centimeters does it hold after 10 days? Since the beach ball loses air. if no more money is added or subtracted from the principal? Sam¶s account will have $2000 1. If 1000 dollars is initially put into a savings account. is to isolate the exponential term. Page 62 . You can then isolate the variable on one side of the equation. and we see that we have a logarithm problem that can be methodically solved. Here¶s a simple example to illustrate this process: If 6x = 5 1 0 0 0 . the initial quantity. is not a polynomial because x is raised to a negative power. each consisting of a constant multiplied by a variable raised to a power greater than or equal to zero. So about halfway through the year 2021. Here.This question is like the exponential growth problems we¶ve just seen but with a twist. it will take roughly 20. How would you possibly calculate 51000 anyhow? And how do you solve for x when it¶s the exponent of a number? But by taking the logarithm of each side of the equation. Since logarithms are the power to which you must raise a given number to equal another number. we¶re given the growth rate. The general form for a problem like this one. The rest of this chapter will show you how to perform different operations on and with polynomials. A binomial is a polynomial with exactly two terms: and are both binomials. In this case. and utilizing the power rule of logarithms: The confusion clears. as long as it is consistently used. You could choose a base-10 logarithm or a logarithm of any other base. For example. and the ending quantity.5 years for the town¶s population to exceed 1500. in which the exponent is unknown. the number of years) that links all these values. This problem would be vastly more difficult if we didn¶t have logarithms. take the logarithm of both sides. on the other hand. the population will first exceed 1500. The base of the logarithms is insignificant. and then use the power rule of logarithms to bring the variable out of the exponent. Polynomials A polynomial is an expression that contains one or more algebraic terms. they are the perfect tool for solving this sort of problem. We need to find the number of percent changes (in this case. then find the value of x. . is a polynomial with three terms (the third term is . Consider the polynomials (a + b + c) and (d + e + f). It may seem like a daunting task. multiply the last terms: Combine like terms and you have your product: Here are a few more examples: Multiplying Polynomials Every once in a while. multiply the inner terms: Finally. But when the process is broken down. multiplying polynomials requires nothing more than distribution and combining like terms. multiply the outer terms of the binomials: Then. Inner. To find their product. It is FOIL. This is the order that you multiply the terms of two binomials to get the right product. if asked to multiply the binomials: You first multiply the first terms of each binomial: Next.Multiplying Binomials There is a very simple acronym that is useful in remembering how to multiply binomials. Outer. and it stands for First. Last. the Math IC test will ask you to multiply polynomials. just distribute the terms of the first polynomial into the second polynomial individually and combine like terms to formulate your final answer: Page Here¶s another example: 63 . For example. The quadratic formula takes longer to work out. since x + 3 = 0 or x + 7 = 0. since 21 is the product of the two last terms of the binomials. Thus. The values of x for which the equation holds are called the roots. Quadratics with Negative Terms So far we¶ve dealt only with quadratics in which the terms are all positive. The only difference between a quadratic equation and a quadratic expression is that the equation is set equal to 0 (x2 + 10x + 21 = 0). the product of m and n is ±21. You also know that the sum of m and n is 10. On the Math IC. a quadratic equation is an equation of the form ax2 + bx + c = 0. Factoring To factor a quadratic. Finally. since the first term of the quadratic is x2. or solutions. Now you just need to put the pieces together to find the values of m and n. Take a look at this quadratic: In the example above. the leading term has a coefficient of 1 (since 1x2 is the same as x2). of the quadratic equation. we know that the binomials whose product is this quadratic must be of the form (x + m)(x + n). you know that the product of m and n equals 21. The pair of numbers that fit the bill for m and n are 3 and 7. and you know that the sum of m and n is 10 and the product of m and n is 21. is a polynomial of the form ax2 + bx + c. The equation also tells you that either m or n must Page 64 . That is. though. There are two basic ways to find roots: by factoring and by using the quadratic fo-rmula. the solutions (also known as the roots) of the quadratic must be x = ±3 and x = ±7. There are a number of things you can tell from this equation: the first term of each binomial is x. where a  0. but it works for all quadratic equations. We¶ll study both in detail. Quadratic Equations A quadratic. factoring a quadratic involves a reverse-FOIL process. then once you have factored the quadratic you can solve it. one of the terms must be equal to zero.As you can see. Factoring a quadratic that has negative terms is no more difficult. but it can¶t always be done. x2 + 10x + 21 = (x + 3)(x + 7). so m + n must equal 10). Most of the questions on quadratic equations involve finding their roots. you will often be presented with a quadratic equation. Since the two x variables are multiplied together during the FIRST step of foiling to get the first term of the quadratic polynomial. Thus. If you have such an equation. The quadratic expression has now been factored and simplified. The following polynomials are quadratics: A quadratic equation sets a quadratic polynomial equal to zero. simply because you are less used to thinking about negative numbers. where m and n are constants. You know that x is the first term of both binomials. Consider the quadratic equation x2 ± 4x ± 21 = 0. Because the product of two terms is zero. Factoring is faster. since the 10x is derived from multiplying the OUTER and INNER terms of the binomials and then adding the resulting terms together (10x = mx + nx. but it might take slightly longer to get the hang of it. and the sum of a and b equals ±4. multiplying polynomials is little more than rote multiplication and addition. In essence. or quadratic polynomial. you must express it as the product of two binomials. To solve this problem by working out the math.be negative but that both cannot be negative. The pair that works in the equation is ±7 and 3. They are: 1. practice. you could immediately have divided out the 2 from both sides of the equation and seen that the solution to the problem is ±5. The Quadratic Formula Page 65 . Example: a2 ± 6ab + 9 = (a ±3)2 Note that when you solve for the roots of a perfect square quadratic equation. There are two kinds of perfect square quadratics. while the solution for (a + b)2 = 0 will be b. Take a look at the following examples and try to factor them on your own before you peek at the answers. They are the perfect square and the difference of two squares. the solution for the equation (a + b)2 = 0 will be ±b. Two Special Quadratic Polynomials There are two special quadratic polynomials that pop up quite frequently on the Math IC. you would do the following: If you got to the step where you had 2(x2 + 10x +25) = 0 and realized that you were working with a perfect square of 2(x + 5)2. If you memorize the formulas below. Example: a2 + 6ab + 9 = (a + 3)2 2. The numbers that multiply together to give you ±21 are: ±21 and 1. and you should memorize them. Now you need to look for the numbers that fit these requirements for m and n. because the multiplication of one positive and one negative number can only result in a negative number. practice. 3 and ±7. you may be able to avoid the time taken by factoring. and 21 and ±1. The difference of two squares quadratics follow the form below: Here¶s an instance where knowing the perfect square or difference of two square equations can help you: Solve for x: 2x 2 + 20x + 50 = 0. Just like perfecting a jump shot. a2 + 2ab + b2 = (a + b)(a + b) = (a + b)2. repeating the same drill over and over again will make you faster and more accurate. ±7 and 3. Practice Quadratics Since the ability to factor quadratics relies in large part on your ability to ³read´ the information in the quadratic. the best way to get good is to practice. a2 ± 2ab + b2 = (a ± b)(a ± b) = (a ± b)2. If: 1. The discriminant of a quadratic is the quantity b2 ± 4ac. 2. all you need to find is an equation¶s discriminant. FINDING THE DISCRIMINANT: If you want to find out quickly how many roots an equation has without calculating the entire formula. b2 ± 4ac< 0. a = 1. say you¶re trying to solve for the speed of a train in a rate problem. the quadratic formula states: Consider the quadratic equation x2 + 5x + 3 = 0. Equations like these can be solved using the quadratic formula. and we must resort to the quadratic equation. b2 ± 4ac> 0. b = 5. and you find that the discriminant is less than zero. and c = 3 into the formula: The roots of the quadratic are approximately {±4. This information is useful when deciding whether to crank out the quadratic formula on an equation. As you can see. We plug the values. and it can spare you some unnecessary computation. b2 ± 4ac = 0. So. and two complex roots.303. For example. can have decimal numbers or fractions as roots. the quadratic has no real roots.697}.Factoring using the reverse-FOIL method is really only practical when the roots are integers. 3. the quadratic has two real roots. and there is no reason to carry out the quadratic formula. This means that there are no real roots (a train can only travel at speeds that are real numbers). however. ±. There are no integers with a sum of 5 and product of 3. For an equation of the form ax2 + bx + c = 0. this quadratic can¶t be factored. this is the radicand in the quadratic equation. Key Formulas Distributive Property Perfect Square of a Binomial Page 66 Difference of Two Squares . Quadratics. the quadratic has one real root and is a perfect square. (A) (B) (C) x={ (D) (D) x={ 4. 8} x = {6. 8} x< ±13 x • ±13 x ” ±5 x< ±5 x> ±5 Solve for x in the following equation: 3 + |2x ± 7| = x + 2 x = ±6 x=8 >3 x = ±3 x=0 x=6 x = 12 x = 24 If x = 2y ± 4 and y = 7s + 1. Solve for x in the following inequality: (A) (B) (C) (D) (D) 3. what is the value of s in terms of x? Page s = 67 (C) . (A) s = (B) s = .Quadratic Formula In a quadratic equation of the form ax2 + bx + c = 0. 6} . Solve for x in the equation (A) (B) (C) (D) (D) 2. where a  0: Review Questions 1. (D) s = (D) s = 5. A second train leaves a different station an hour later.5 25 8.20 $13.30 $14. How many full days will pass before it weighs less than 100 pounds? (A) (B) (C) (D) (D) 17 22 23 27 45 What is (a + b + 3c)(2a + 3b + 4c)? 9. If Jim runs h laps per hour. Jim and Ryan run laps around a track which is 1 / 4 of a mile long.20 $11. A snowman weighing 250 pounds begins to melt in the spring. how many miles farther than Ryan has Jim run two hours after Jim starts? (A) (B) (C) (D) (D) +2+ 6. If the stations are 255 miles apart. traveling due west on the same track. Ryan runs half as fast as Jim.20 A train leaves the station traveling due east at a rate of 45 miles per hour. how many miles from the halfway point between the stations will the trains collide? (A) (B) (C) (D) (D) 0 5 7. He then resells it at a 60% increase from the price at which he bought it. It loses 4% of its weight every day. Page 68 . Ken buys a shirt on a 30% sale from its original price of 10 dollars. going 60 miles per hour. How much does Ken sell the shirt for? $4. and Jim starts running half an hour before Ryan.5 12. (A) (B) (C) (D) (D) 7.00 $13. D To solve this problem. D Solving for a variable in an inequality is similar to solving for a variable in a normal equation.(A) (B) (C) (D) (D) 10. you must reverse the direction of the inequality symbol: 3. In this first case. 9} x = {3. the expression within absolute value brackets is positive: . ±9} x = {±1. just isolate the variable: 2. E First. But remember that if you multiply or divide an inequality by a negative number. with one big difference: you isolate the variable. (A) (B) (C) (D) (D) 3a + 4b + 7c 2a 2 + 3b 2 + 12c 2 2a 2 + 10ac + 12c 2 2a 2 + 3ab + 10ac + 9bc + 12c 2 2a 2 + 3b 2 + 12c 2 + 5ab + 10ac + 13bc Solve for x in the following equations: 3x 2 + 24x ± 27 = 0 x=1 x = ±9 x = {1. isolate the expression within the absolute value brackets: Page 69 Then divide the equation into two equations. 27} Explanations 1. 25h»4 miles.75h»4 = 1. so his total distance traveled is 2h laps. and so we perform each percent change one by one. we can plug the known values into the rate formula: Jim runs h laps per hour for 2 hours.5 hours. 6}. B First. Ken bought the shirt at a discount of 30%. substitute this value for y in the equation with s: 5. First.75h»4 miles. let the expression within absolute value brackets be negative: The two solutions are x = {8»3. The difference between the distance traveled by Jim and the distance traveled by Ryan is 2h»4 ± . B This is a double percent-change problem. Ryan runs . and we are given the input and rate in order to find the output. 4. A This is a rate question. This is equal to .Next.75h laps. solve for y in terms of x: Finally. which equals 2h»4 miles. so in total he runs 0. 6.5h laps per hour for 1. The price at which he paid was: He then sold it for 60% more than he paid: Page 70 . So. solve for s in terms of y: Next. He sold the shirt for $11. Here is the rate formula we¶ll be using: Let x represent the number of hours before the trains collide. 8. and 2 60 = 120 miles from the eastern station. From this. E To multiply polynomials two at a time. as the question asked. we know that the collision happened 3 45 = 135 miles from the western station. Their combined distance traveled is 255. from this newly calculated information. C This problem fits the classic exponential decay model.5 miles from the halfway point between the stations. Finally. The point at which the trains will collide is the point at which their combined distance traveled is 255 miles. in relation to when the trains left their respective stations. we can find out when the collision occurs. so it happened 135 ± 127. or. C The toughest part of this rate problem is translating the word problem into an equation. 7.5 miles from either station.5 days to reach the 100 pound mark. Using this fact and the rates at which the trains travel. we can find where the collision occurred. We then have the equation: This equation explains the situation before the collision: that the train going 45 miles per hour traveled for x hours and the train traveling 60 miles per hour traveled for x ± 1 hours. just distribute the terms of one polynomial into the other one individually: Page 71 .5 = 7. 23 full days.20. it takes approximately 22. So we plug the given information into the formula: Then we solve: Thus. Now solve the equation for x: 3 hours pass before the trains collide. 9. The halfway point between the stations is 255»2 = 127. Plane Geometry ROUGHLY 20 PERCENT OF THE QUESTIONS on the Math IC test cover plane geometry. a whopping 45 percent of the Math IC test involves plane geometry either directly or indirectly. Distance and Midpoint of a Line Segment The midpoint of a line segment is the point on the segment that is equidistant (the same distance) from each endpoint. so you can use either the reverse FOIL or the quadratic formula to solve for the roots. and you get: The solution set for x is {1. first factor out 3 from the equation: Factoring takes less time than working out the quadratic formula. The Math IC writers usually make their questions a little trickier though. Because a midpoint cuts a line segment in half. questions that deal more specifically with solid geometry. For example. coordinate geometry. a line segment has finite length. The Math IC test often asks questions that focus on this property of midpoints.10. Take a look: Page 72 . Lines and Angles A line is a collection of points that extends without limit in a straight formation. by including multiple midpoints. the length of the whole line segment is 10. the midpoint is said to bisect the line segment. Line segment AB is pictured below. like line AB. ±9}. Unlike a line. Because a midpoint splits a line segment into two equal halves. so check to see if factoring is possible. Before doing either of those things. J and K: a line is determined: This line is called JK. if the distance from one endpoint to the midpoint of a line segment is 5. It is. For example. The second way of naming a line indicates an important property common to all lines: any two points in space determine a line. C The equation given is in the form of a quadratic equation ax2 + bx + c = 0. given two points. A line can be named by a single letter. or it can be named according to two points that it contains. knowing the distance between the midpoint and one endpoint of a line segment allows you to calculate the length of the entire line segment. It is named and determined by its endpoints. Line Segments A line segment is a section of a line. whose length is infinite. and trigonometry assume a thorough knowledge of plane geometry. In all. like line l. In addition. Angles Technically speaking. The measure of an angle is how far you must rotate one of the rays such that it coincides with the other. is 90 . you know that XY = YZ and that both XY and YZ are equal to 1» XZ and 1» WZ. you know that XM = MY and that both XM and MY are equal to 1»2XY and 1»8WZ. The question asks for the length of WX. MY = 3. so WX = 12.X is the midpoint of WZ and Y is the midpoint of XZ. you don¶t really need to bother with such a technical definition. Please note that you don¶t have to write out these relationships when answering this sort of question. it¶s possible to see the relationships. you know that WX = XZ and that both WX and XY are equal to 1»2WZ. and the new line segments that the midpoints create. you can solve the problem. all relate to each other. 2 4 Since M is the midpoint of XY. Vertical Angles When two lines or line segments intersect. Instead of trying to visualize what is being described in your head. you can see how the three midpoints. draw a sketch of what the question describes. y y y Since X is the midpoint of WZ. Suffice it to say. also known as a straight angle. which is equal to 1»2WZ. Once you know the relationships. what is the length of WX? All the midpoints flying around in this question can get quite confusing. The angles in each pair of congruent angles created by the intersection of two lines are called vertical angles: Page 73 . as the question tells you. is 180 degrees. For this question. Since Y is the midpoint of XZ. If you draw a good sketch. you know that MY is equal to 1»8WZ. A half-revolution. angles can also be indicated by the symbol . an angle is the union of two rays (lines that extend infinitely in just one direction) that share an endpoint (called the vertex of the angle). One full revolution around a point creates an angle of 360 degrees. Since. or right angle. A quarter revolution. Once you¶ve drawn a sketch. or 360 . two pairs of congruent (equal) angles are created. If M is the midpoint of XY and MY = 3. In text. In this guide and for the Math IC. you can calculate that WZ = 24. angles are used to measure rotation. and . you also have supplementary angles. In the figure above. and and . The intersection of one line with two parallel lines creates many interesting angle relationships. and . and are supplementary. In the figure above. and . and and . Whenever you have vertical angles. Alternate exterior angles are pairs of congruent angles on opposite sides of the transversal. and are vertical angles (and therefore congruent). 3. In the previous figure.In this figure. and . Parallel Lines Cut by a Transversal Lines that will never intersect are called parallel lines. Supplementary and Complementary Angles Supplementary angles are two angles that together add up to 180º. Complementary angles are two angles that add up to 90º. while the other will be outside the parallel lines. In the diagram of vertical angles above. all adjacent angles formed when two parallel lines are cut by a transversal are supplementary. as are and . and and are all pairs of supplementary angles. one will always be between the parallel lines.´ where the transversal is the nonparallel line. In addition to these special relationships between angles. As you can see in the diagram below of parallel lines AB and CD and transversal EF. and . three special angle relationships exist: 1. This situation is often referred to as ³parallel lines cut by a transversal. and and . Corresponding angles are congruent angles on the same side of the transversal. there are four pairs of corresponding angles: and . Of two corresponding angles. Math IC questions covering parallel lines cut by a transversal are usually straightforward. In the figure above. Alternate interior angles are pairs of congruent angles on opposite sides of the transversal in the region between the parallel lines. which are given by the symbol ||. two parallel lines cut by a transversal will form eight angles. 2. Among the eight angles formed. outside of the space between the parallel lines. there are two pairs of alternate exterior angles: and . there are two pairs of alternate interior angles: and . For example: Page 74 . for example. From here. since AD = DB. answering this question is easy. then those two lines or line segments are parallel. Keep in mind that if a single line or line segment is perpendicular to two different lines or line segments. you will be able to find the third. if you know the measures of two of a triangle¶s angles.In the figure below. it¶s easy to calculate that f ± g = 110º ± 70º = 40º. Luckily for you. Perpendicular Lines Two lines that intersect to form a right (90º) angle are called perpendicular lines. so it must be equal to 180º ± 110º = 70º. outlined below: 1. forming vertical angles of 90º in the process. A line or line segment is called a perpendicular bisector when it intersects a line segment at the midpoint. With this rule. Line segments AB and CD are perpendicular. Helpful. don¶t you think? Page 75 . and trigonometry. Not only will you encounter numerous questions specifically about triangles. you will also need a solid understanding of triangles in order to answer other questions about polygons. CD is the perpendicular bisector of AB. coordinate geometry. the essential rules governing triangles are few and simple to master. but it is a common situation when dealing with polygons. and are alternate exterior angles. is adjacent to . Basic Properties Every triangle adheres to four main rules. in the figure above. This is actually just another example of parallel lines being cut by a transversal (in this case. Triangles The importance of triangles to the plane geometry questions on the Math IC test cannot be overstated. Sum of the Interior Angles If you were trapped on a desert island and had to take the Math IC test. the transversal is perpendicular to the parallel lines). then f ± g = If you know the relationships of the angles formed by two parallel lines cut by a transversal. if lines m and n are parallel and = 110º. so . this is the one rule about triangles you should bring along: the sum of the measures of the interior angles is 180º. For example. We¶ll examine this type of case later. If this angle is small (close to 0º). Take a look at the figure below. Observe the figure below: From the triangle inequality. Page 76 . the exterior angle will always be supplementary to the interior angle with which it shares a vertex and therefore (because of the 180º rule) equal to the sum of the remote interior angles. which states: the length of a side of a triangle is less than the sum of the lengths of the other two sides and greater than the difference of the lengths of the other two sides. In every triangle. which has to do with the relationships between the angles of a triangle and the lengths of the triangle¶s sides. and equal in measure to the sum of the measures of the remote interior angles. we can tell that 9 ± 4 <x< 9 + 4. If this angle is large (close to 180º) then a will be large (close to b + c). we know that c ± b < a < c + b. Measure of an Exterior Angle Another property of triangles is that the measure of an exterior angle of a triangle is equal to the sum of the measures of the remote interior angles. then a will be small (close to b ± c). take a look at this triangle: Using the triangle inequality. Proportionality of Triangles This brings us to the last basic property of triangles. The exact length of side a depends on the measure of the angle created by sides b and c.2. An exterior angle is always supplementary to the interior angle with which it shares a vertex. in which d. The exact value of x depends on the measure of the angle opposite side x. the exterior angle. is supplementary to interior angle c: It doesn¶t matter which side of a triangle you extend to create an exterior angle. or 5 <x< 13. 4. the longest side is opposite the largest angle and the shortest side is opposite the smallest angle. 3. An exterior angle of a triangle is the angle formed by extending one of the sides of the triangle past a vertex (the point at which two sides meet). For an example. Triangle Inequality The third important property of triangles is the triangle inequality rule. side a is clearly the longest side and is the largest angle. Conversely. which is opposite the side of length 8. we could find the measure of the third angle and therefore decide which side is the longest (it would be the side opposite the largest angle). if we knew the measures of two of the angles in the triangle. The value 7 is the only choice that fits the criteria. It follows.In this figure. We did not assign measures and lengths to the angles and sides for the figure above. measures between 6 and 8 units in length. on to the special triangles. This is the kind of reasoning that you might have to use when dealing with triangles on the test. If we had limited information about those values. which is opposite the side of length 6. we must take a moment to explain the markings we use to describe the properties of each particular triangle. What is one possible value of x if angle C < A < B? (A) (B) (C) (D) (D) 1 6 7 10 15 The largest angle in triangle ABC is . The arcs drawn into and indicate that these angles are congruent. In some diagrams. Knowing these triangles and what makes each of them special will help you immeasurably on the Math IC test. For example. of length x. Page 77 . But before getting into the different types of special triangles. that c < b < a and C < B < A. double hash marks or double arcs can be drawn into a pair of sides or angles to indicate that they are equal to each other. For example. Special Triangles There are several special triangles that have particular properties. the figure below has two pairs of sides of equal length and three congruent angle pairs: these indicate that the sides have equal length. but not necessarily equal to the other pair of sides or angles: Now. therefore. This means that the third side. there might be more than one pair of equal sides or congruent angles. The smallest angle in triangle ABC is . however we could make certain assumptions about the other unknown side lengths and angles measures. side c is the shortest side and is the smallest angle. In this case. This proportionality of side lengths and angle measures holds true for all triangles. Scalene Triangles A scalene triangle has no equal sides and no equal angles. In an isosceles triangle. In the isosceles triangle below. the two angles opposite the sides of equal length are congruent. and sides a and b are the legs. the special property of scalene triangles is that they don¶t really have any special qualities. For example. the measure of each is 60º. then you know it has three equal sides. Right Triangles A triangle that contains a right angle is called a right triangle. is the right angle (as indicated by the box drawn in the angle). Scalene triangles almost never appear on the Math IC.In fact. Since the three angles in a triangle must add up to 180º. The side opposite the right angle is called the hypotenuse of the right triangle. All three angles in an equilateral triangle are congruent as well. you can figure out the value of the third angle: 180º ± 35º ± 35º = 110º. These angles are usually referred to as base angles. The angles opposite the legs of a right triangle are complementary. side c is the hypotenuse. If you know that a triangle has three equal sides. then the proportionality rule states that the triangle must also have three equal angles. Isosceles Triangles A triangle that contains two sides of equal length is called an isosceles triangle. Equilateral Triangles A triangle whose sides are all of equal length is called an equilateral triangle. side a = b and : If you know the value of one of the base angles in an isosceles triangle. and the other two sides are called legs. you can figure out all the angles. . The Pythagorean Theorem Page 78 In the figure above. if one base angle of an isosceles triangle is 35º. if you know that a triangle has three equal angles. then you know that the other base angle is also 35º. Similarly. . 15. For example. has length x. For example. If. 12. opposite the 30 degree angle. and 90º. 4. The theorem states that in a right triangle a2 + b2 = c2: where c is the length of the hypotenuse. 60º. 8. PYTHAGOREAN TRIPLES Because right triangles obey the Pythagorean theorem. 30-60-90 Triangles A 30-60-90 triangle is a triangle with angles of 30º. Then the hypotenuse has length 2x. 17} In addition to these Pythagorean triples. since it is a multiple of {3. 13} {7. 5}. What makes it special is the specific pattern that the lengths of the sides of a 30-60-90 triangle follow. Study the ones above and their multiples. you can drastically reduce the amount of time you need to spend on the problem since you won¶t need to do any calculations. 10} is a Pythagorean triple. 5} {5. for Page 79 . The Math IC is full of right triangles whose side lengths are Pythagorean triples. has length x ratio of 1 : 2 : . and a and b are the lengths of the two legs. The few sets of three integers that do obey the Pythagorean theorem and can therefore be the lengths of the sides of a right triangle are called Pythagorean triples. If you know the measures of two sides of a right triangle. a right triangle with legs of length 3 and 5 has a hypotenuse of length = 5. The sides of every 30-60-90 triangle will follow this The constant ratio of the lengths of the sides of a 30-60-90 triangle means that if you know the length of one side in the triangle. Here are some common ones: {3. 24. you will immediately know the lengths of all the sides.83. Special Right Triangles Right triangles are pretty special in their own right. But there are two extraspecial right triangles that appear frequently on the Math IC. 4. 25} {8. Suppose the short leg. The theorem states that the square of the hypotenuse is equal to the sum of the squares of the two legs. They are 30-60-90 triangles and 45-45-90 triangles. as you study coordinate geometry and trigonometry. If you can recognize a Pythagorean triple on a triangle during the test. you can always use the Pythagorean theorem to find the third. opposite the 60 degree angle. {6.The Pythagorean theorem is crucial to answering most of the right-triangle questions that you¶ll encounter on the Math IC. The theorem will also come in handy later on. and the long leg. you should also watch out for their multiples. only a few have side lengths which are all integers. considering that ABC ~ DEF. When you say that two triangles are similar. 80 .example. you know that the ratio of the short sides equal the ratio of the larger sides. the symbol for ³is similar to´ is ~. Just as similar triangles have corresponding sides. the Math IC may present you with two separate triangles and tell you that the two are similar. and the leg opposite the 60º angle is 2 meters. smaller. You could solve these questions by using the Pythagorean theorem. the lengths of the sides of a 45-45-90 triangle also follow a specific pattern that you should know. This line segment creates a second. similar triangles have exactly the same shape but not necessarily the same size. and triangle ABC is similar to triangle AED. then the hypotenuse has length x Take a look at this diagram: . In essence. On the Math IC. Take a look at a few similar triangles: Page As you may have assumed from the above figure. AB/DE = BC/EF = CA/FD. for example. As with 30-60-90 triangles. line segment DE is parallel to CB. the definition of similar triangles is that ³the ratio of the lengths of their corresponding sides is constant. then by using the 1 : 2 : ratio. So if triangle ABC is similar to triangle DEF. but that method takes a lot longer than simply knowing the proper 30-60-90 ratio. If the legs are of length x (they are always equal). knowing the 1: 1: deal of time on the Math IC. Like the 3060-90 triangle. If ABC ~ DEF. since it¶s both isosceles and right. More often. ratio for 45-45-90 triangles can save you a great Similar Triangles Two triangles are called similar if the ratio of the lengths of their corresponding sides is constant. it is important to know which sides of each triangle correspond to each other. allowing you to use your knowledge of this special triangle.´ So. After all. they also have corresponding angles. similar triangle. you will write ABC ~ DEF. you will quite often encounter a question that will present you with an unnamed 30-60-90 triangle. the corresponding angles of each triangle must be congruent. This type of triangle is also sometimes referred to as an isosceles right triangle. you will know that the hypotenuse is 4 meters long. In the figure below. In order for this to be true. you know that the side opposite the 30º angle is 2 meters long. the Math IC will present you with a single triangle that contains a line segment parallel to one base. 45-45-90 Triangles A 45-45-90 triangle is a triangle with two 45º angles and one right angle. then Occasionally. you know that the corresponding sides of the two triangles are in constant proportion.After presenting you with a diagram like the one above. Study the triangle on the right. xº. To solve for DE. Once you realize that ABC ~ AED. This is why the altitude of a triangle is defined as a line segment perpendicular to the line containing the base and Page 81 . The height of the triangle depends on the base. and h is height (also called the altitude). Its altitude does not lie in the interior of the triangle. In the previous sentence we said ³a base´ instead of ³the base´ because you can actually use any of the three sides of the triangle as the base. no matter which side you choose to be the base. when they intersect with BA. which is why the area formula always works. the Math IC will test whether you understand similarity by asking a question like: If = 6 and = . but it implicitly tells you that the two lines are parallel by indicating that both lines form the same angle. you have to plug it into the proportion along with CB: Area of a Triangle It¶s quite likely that you will have to calculate the area of a triangle for the Math IC. a triangle has no particular side that is the base until you designate one. what is ? This question doesn¶t tell you outright that DE and CB are parallel. The formula for the area of a triangle is: where b is the length of a base of the triangle. The question tells you what this proportion is when it tells you that AD = 2»3 AC. The heights of a few triangles are pictured with their altitudes drawn in as dotted lines. Sometimes the endpoint of the altitude does not lie on the base. It¶s more probable that the altitude would have to be found. draw in the altitude from the base (of length 9) to the opposite vertex. The perimeter of the hexagon below. you may be tested on the area of a triangle in a few different ways. share certain characteristics: y y y Page The sum of the interior angles of a polygon with n sides is (n ± 2) . but it¶s unlikely you¶d get such an easy question. is 35. triangles are actually a type of polygon. for example. The perimeter of a polygon is the sum of the lengths of its sides. no matter the number of sides they possess. but they are so important on the Math IIC that we gave them their own section. For example. So. )=9 Polygons Polygons are enclosed geometric shapes that cannot have fewer than three sides. As this definition suggests. All polygons. Notice that now you have two triangles. for example. The sum of the exterior angles of any polygon is . as you can see in the chart below. On the Math IC. Polygons are named according to the number of sides they have. it can be outside of the triangle. which is also the altitude of the original triangle. the sum of the interior angles of an octagon is (8 ± 2) =6 = . is 2 the original triangle: 2bh = 2(9)(2 1» 1» . You might be given the altitude of a triangle along with the length of the base. Now you can use the area formula to find the area of § 15. 82 . try to find the area of the triangle below: To find the area of this triangle. The hypotenuse of this 30-60-90 triangle is 4. so the short side is 2 and the medium side. as is the case of the second example above.6.not simply as perpendicular to the base. and one of them (the smaller one on the right) is a 30-60-90 triangle. using other tools and techniques from plane geometry. of a regular pentagon. as shown in the sections below. A regular hexagon can be divided into six equilateral triangles. rhombuses. It has to do specifically with regular hexagons. Quadrilaterals The most frequently seen polygon on the Math IC is the quadrilateral. which is a general term for a four-sided polygon. Below are diagrams. The area of a trapezoid is: Page 83 . parallelograms. you can use that information to calculate the area of the equilateral triangle that uses the side. there are five types of quadrilaterals that pop up on the test: trapezoids. and a square (also known as a regular quadrilateral): Area of a Regular Polygon There is one more characteristic of polygons with which to become familiar. In fact. Each of these five quadrilaterals has special qualities. and squares. To find the area of the hexagon. a regular octagon. AB is parallel to CD (shown by the arrow marks). simply multiply the area of that triangle by 6.Regular Polygons Most of the polygons with more than four sides that you¶ll deal with on the Math IIC will be regular polygons²polygons whose sides are all of equal length and whose angles are all congruent (neither of these conditions can exist without the other). Trapezoids A trapezoid is a quadrilateral with one pair of parallel sides and one pair of nonparallel sides. from left to right. Below is an example of a trapezoid: In the trapezoid pictured above. whereas AC and BD are not parallel. rectangles. as the figure below shows: If you know the length of just one side of a regular hexagon. Check out the figure below. Opposite sides are equal. 2. Opposite angles are congruent. The figure below shows an example: Parallelograms have three very important properties: 1. Adjacent angles are supplementary (they add up to 180º). draw in the height of the trapezoid so that you create a 45-45-90 triangle. Try to find the area of the trapezoid pictured below: To find the area. You know that the length of the leg of this triangle²and the height of the trapezoid²is 4. which includes all the information we know about the trapezoid: Parallelogram A parallelogram is a quadrilateral whose opposite sides are parallel. and h is the height. and h is the height. 3. the height is the perpendicular distance from one base to the other. simply picture the opposite sides of the parallelogram as parallel lines and one of the other sides as a transversal: The area of a parallelogram is given by the formula: Page 84 where b is the length of the base.where s1 and s2 are the lengths of the parallel sides (also called the bases of the trapezoid). the area of the trapezoid is 6+10»2 4 = 8 4 = 32. Thus. In a trapezoid. . To visualize this last property. you can always calculate the third with the Pythagorean theorem. The formula for the area of a rectangle is: where b is the length of the base.In area problems. In the figure below. Because the diagonal of the rectangle forms right triangles that include the diagonal and two sides of the rectangle. rhombuses. A rectangle is essentially a parallelogram in which the angles are all right angles. and squares²are all special types of parallelograms. and h is the height. the diagonal BD cuts rectangle ABCD into congruent right triangles ABD and BCD. Also similar to parallelograms. if you know the diagonal and one side length. and h is the height. Page 85 . you can calculate the diagonal. A diagonal through the rectangle cuts the rectangle into two equal right triangles. if you know two of these values. the opposite sides of a rectangle are equal. you will likely have to find the height using techniques similar to the one used in the previous example problem with trapezoids. The next three quadrilaterals that we¶ll review²rectangles. Rhombuses A rhombus is a quadrilateral in which the opposite sides are parallel and the sides are of equal length. The formula for the area of a rhombus is: where b is the length of the base. you can calculate the other side. Rectangles A rectangle is a quadrilateral in which the opposite sides are parallel and the interior angles are all right angles. If you know the side lengths of the rectangle. and test questions will focus on your understanding of these properties. and you can calculate the length of this altitude to be 2 =8 . The distance from the center to any point on the circle is called the radius. Unlike polygons. called the center. it is also possible to provide a simple formula for the perimeter: P = 4s. where s is. rectangle. and ABD is an equilateral triangle. so the area of this rhombus is 4 2 Squares A square is a quadrilateral in which all the sides are equal and all the angles are right angles. and parallelogram: The formula for the area of a square is: where s is the length of a side of the square. if you know the length of one side of the square. what is the area of the rhombus? If ABD is an equilateral triangle. the length of a side. Similarly. The area of a rhombus is bh. If you know the radius of a circle. A circle is named after its center point. you can easily calculate the length of the diagonal. . then the length of a side of the rhombus is 4. AC = 4. Circles have certain basic characteristics.To find the area of a rhombus. (r). once again. use the same methods as used to find the area of a parallelogram. For example: If ABCD is a rhombus. you can calculate the length of the sides of the square. Draw an altitude from a to DC to create a 30-60-90 triangle. if you know the length of the diagonal. Circles Circles are another popular plane-geometry test topic. which is the most important measurement in a circle. A circle is the collection of all points equidistant from a given point. we know that . A diagonal drawn into the square will always form two congruent 45-45-90 triangles: From the properties of a 45-45-90 triangle. and angles ADB and ABD are 60º. all circles are the same shape and vary only in size. Page 86 Basic Definitions of Circles .In other words. A square is a special type of rhombus. Because all the sides of a square are equal. r is the radius. and AB is a chord. The Math IC often includes tangent lines in the test. XY = XZ. point C is the center of the circle.you can figure out all its other characteristics. but it does not necessarily pass through the center. In the figure below. The distance from the origin of the two tangents to the points of tangency are always equal. and stretches between endpoints on the circle. The diameter (d) of a circle is twice as long as the radius (d = 2r). Every point in space outside the circle can extend exactly two tangent lines to the circle. In the figure below. making sure to pass through the center. if RS is tangent to circle Q? Page 87 . A chord also extends from endpoint to endpoint on the circle. Tangent Lines A line that intersects the circle at only one point is called a tangent line. For example: What is the area of triangle QRS. The radius whose endpoint is the intersection point of the tangent line and the circle is always perpendicular to the tangent line. If RS is tangent to circle Q, then QR is perpendicular to RS, and therefore QRS is a 30-60-90 triangle. Given that QR = 4, we know that RS = 4 . , and the area of triangle QRS is 1»2(4)(4 )=8 Central Angles and Inscribed Angles An angle whose vertex is the center of the circle is called a centralangle. The degree of the circle (the slice of pie) cut by a central angle is equal to the measure of the angle. If a central angle is 25º, then it cuts a 25º arc in the circle. An inscribedangle is an angle formed by two chords in a circle that originate from a single point. An inscribed angle will always cut out an arc in the circle that is twice the size of the degree of the inscribed angle. If an inscribed angle has a degree of 40º, it will cut an arc of 80º in the circle. If an inscribed angle and a central angle cut out the same arc in a circle, the central angle will be twice as large as the inscribed angle. Circumference of a Circle Page 88 The circumference of a circle is the length of the 360º arc that forms the circle. In other words, if you were to trace around the edge of the circle, it is the distance from a point on the circle back to itself. The circumference is the perimeter of the circle. The formula for circumference is: where r is the radius. The formula can also be written C = d, where d is the diameter. Using the formula, try to find the circumference of the circle below: Plugging the radius into the formula, C = 2 r = 2 (3) = 6 . Arc Length An arc is part of a circle¶s circumference. An arc contains two endpoints and all the points on the circle between the endpoints. By picking any two points on a circle, two arcs are created: a major arc, which is by definition the longer arc, and a minor arc, which is the shorter one. Since the degree of an arc is defined by the central or inscribed angle that intercepts the arc¶s endpoints, you need only know the measure of either of those angles and the measure of the radius of the circle to calculate the arc length. The arc length formula is: where n is the measure of the degree of the arc, and r is the radius. The formula could be rewritten as arc length = n»360 C, where C is the circumference of the circle. A Math IC question might ask: Circle D has radius 9. What is the length of arc AB? Page In order to figure out the length of arc AB, you need to know the radius of the circle and the measure of , which is the inscribed angle that intercepts the endpoints of AB. The question tells you the radius of the circle, but it throws you a little curveball by not providing you with the measure of . Instead, the question puts in a triangle and tells you the measures of the other two angles in the triangle. Using this information you can figure out the measure of . Since the three angles of a triangle must add up to 180º, you know that: 89 Since angle c is an inscribed angle, arc AB must be 120º. Now you can plug these values into the formula for arc length Area of a Circle The area of a circle depends on the radius of the circle. The formula for area is: where r is the radius. If you know the radius, you can always find the area. Area of a Sector A sector of a circle is the area enclosed by a central angle and the circle itself. It¶s shaped like a slice of pizza. The shaded region in the figure below is a sector: The area of a sector is related to the area of a circle the same way that the length of an arc is related to circumference. To find the area of a sector, simply find what fraction of 360º the sector comprises and multiply this fraction by the area of the circle. where n is the measure of the central angle which forms the boundary of the sector, and r is the radius. Try to find the area of the sector in the figure below: Page 90 The sector is bounded by a 70º central angle in a circle whose radius is 6. Using the formula, the area of the sector is: Finding the radius of the circle is a little tougher. we can use the arc length formula to find the length of major arc BE. figuring out the measure of the central angle is simple. and the length is 6. where b is the length of the base and h is height.Polygons and Circles You could potentially see a question or two on the Math IC that involve polygons and circles in the same figure. Area = bh. and you know the values of both a and b. Area of a Trapezoid Area = h. Rectangle. where s1 and s2 are the lengths of the bases of the trapezoid. and rectangles only have right angles. you can use the fact that the area of the rectangle is 18. Here¶s an example: What is the length of major arc BE if the area of rectangle ABCD is 18? To find the length of major arc BE. where a and b are the lengths of the legs of a right triangle. To find the height of the rectangle. Page Area of a Parallelogram. you must know two things: the measure of the central angle that intersects the circle at its endpoints and the radius of the circle. Because ABCD is a rectangle. From the diagram. you can see that it is equal to the height of the rectangle. Since A = bh. and c is the length of the hypotenuse. and h is the height. where n is the number of sides in the polygon. Key Formulas Pythagorean Theorem a2 + b2 = c2. and Rhombus 91 . and h is the height. so the measure of the central angle is 360º ± 90º = 270º. With a radius of 3. is 90º. Sum of the Interior Angles of a Polygon The sum of the interior angles of a polygon is (n ± 2)180°. where b is the length of the base. Area of a Triangle Area = bh. (A) (B) (C) (D) (D) 5 5 Page I. where s is the length of a side of the square. lines l and m are parallel. Area of a Sector Area of Sector = r2. and r is the radius of the Area of a Circle Area = r2. Review Questions 1. where n is the measure of the degree of the arc. and AB = 5.Area of a Square Area = s2. where r is the radius of the circle. In the figure below. In the figure below. where r is the radius of the circle. 92 2. and the area of triangle CDB is equal to the area of triangle CEA. line CF is the perpendicular bisector of AB. E and D are the midpoints of BF and AF. where n is the measure of the central angle which forms the boundary of the sector. Circumference of a Circle Circumference = 2 r. and r is the radius of the circle. respectively. 2 r. Which of the following statements must be true? . II. Arc Length Arc Length = circle. What is the perpendicular distance between lines l and m? Note: Figure may not be drawn to scale. Line CF is the perpendicular bisector of ED. Note: Figure may not be drawn to scale. A triangle and a square have the same base and equal areas. If the length of the common base is x and the height of the triangle is h. (A) (B) (C) (D) (D) I only II only III only I and II only I and III only 3. 4x If ABCD is a rhombus and ABD is an equilateral triangle. what is the height of the triangle in terms of x? (A) (B) (C) (D) (D) 4. what is the area of the rhombus? 2x Note: Figure may not be drawn to scale. (A) (B) (C) 4 4 Page 93 . Triangles CEB and CDA have the same area.III. C . the angle whose measure is labeled as 120º is supplementary to Now we have a 30-60-90 triangle whose longer leg. and c? Note: Figure may not be drawn to scale. 8 16 Circle D has radius 8. What is the length of arc AB in terms of a. so its length is 2. (A) (B) (C) (D) (D) Explanations 1.(D) (D) 5. Page 94 The long leg has a : ration to the short leg. . b. is also the distance between lines l and m. Let¶s analyze each statement separately. The short leg has a hypotenuse. so its length is 5 »2. E 5/2. Using the : : : side ratios for 30-60ratio to the 90 triangles you can use the hypotenuse length to calculate the lengths of the other two legs. AC. Because lines l and m are parallel and line AB is a transversal. 2. All we know is that BE = AD. F is the midpoint of AB. then by subtracting DE from each segment. this is given. and all the sides of the rhombus have a length of 4 (by definition of a rhombus. as long as they are equidistant from F. Draw an altitude If ABD is an equilateral triangle. respectively. 3. this statement is not necessarily true. C The area of a triangle with base x and height h is given by the formula 1»2xh. and from the length ratio of x : altitude to be 2 5. So the measure of this angle is 2cº. by definition of a rhombus. we have BE = AD and thus EF = DF. So statement I is true. opposite angles are congruent. This statement is true. Now simply plug the values into the formula: the length of arc AB is: 2 (8) = = Page 95 . Only statements I and III must be true. you can set the two expressions equal to each other and solve for h: The correct answer is h = 2x. D . E and D could be anywhere along BF and AF. If BD = AE. 4. 3. The area of a rhombus is bh. and by definition. Also. we see that triangles CEB and CDA must have the same area. Triangles CDB and CEA are equal in area. . By subtracting the area of triangle CED from each of these triangles. Thus. and since the area of a triangle is found by the formula 1»2b h. It is also given that the area of triangle CDB is equal to the area of triangle CEA. then AD = AB = BD = 4. EF = DF. The area of a square with sides of length x is x2. D x : 2x among the sides. so there is no guarantee that they are the midpoints of BF and AF. These two triangles share the same height. We know that BF = AF because it is given that line CF is the perpendicular bisector of AB. all these conditions hold. so the area of this rhombus is 4 The length of the arc depends on the circumference of the circle and the measure of the central angle that intercepts that arc. and BF = AF. This statement is simply not backed by any evidence. you can calculate the length of this 2 =8 . Angle c is the inscribed angle or one-half as large as the central angle that intercepts the circle at the same points. This statement implies that DF = EF.1. all sides are congruent). it follows that if their areas are equal. As long as points E and D are equidistant from F. Since you know the two shapes have equal areas. The formula is: where n is the measure of the central angle that intercepts the arc and r is the radius. respectively. their bases are equal too. so from a to DC to create a 30-60-90 triangle. The bases of these prisms are shaded. circle. or variations on prisms. we now have cubes and spheres. A prism is defined as a geometric solid with two congruent bases that lie in parallel planes. but there are only a few specific solids that you¶ll need to know about for the Math IC. Prisms Most of the solids you¶ll see on the Math IC test are prisms. or polygon through space without rotating or tilting it. we cover the specifics of calculating surface area as we cover each type of geometric solid. As with volume. and diagonal length. These threedimensional shapes may be more difficult to visualize. Certain geometric solids have slightly different formulas for calculating volume that we will cover on a case-by-case basis. The general formula for calculating the volume of a prism is very simple: where B is the area of the base. The formula for the surface area of a prism therefore depends on the type of prism with which you are dealing. Rectangular Solids A rectangular solid is a prism with a rectangular base and lateral edges that are perpendicular to its base. The figures below are all prisms. Surface Area The surface area of a prism is the sum of the areas of all the prism¶s sides. The perpendicular distance between the bases is the height of the prism. The three-dimensional space defined by the moving triangle or polygon is the body of the prism. width (w). a rectangular solid is shaped like a box. volume. You can create a prism by dragging any two-dimensional triangle. The prism¶s two bases are the planes where the two-dimensional shape begins and ends. you can find the solid¶s surface area. Volume of a Rectangular Solid The volume of a rectangular solid is given by the following formula: Page 96 A rectangular solid has three important dimensions: length (l). If you know these measurements. and the altitude (the height) of each prism is marked by a dashed line: There are two main aspects of geometric solids that are relevant for the Math IC: volume and surface area. We¶ll review them one by one. and height (h). . and h is the prism¶s height.Solid Geometry SOLID GEOMETRY ADDS LITERALLY ANOTHER dimension to the plane geometry explained in the previous chapter²instead of squares and circles. In short. Volume of a Prism The volume of a prism is the amount of space taken up by that prism. The surface area formula is derived by simply adding the areas of the faces²two faces have areas of l w. d. Every rectangular solid has four diagonals. Simply plug the values into the formula given for volume. and h is the height. Surface Area The surface area of a rectangular solid is given by the following formula: where l is the length. try to find the surface area of the rectangular solid we used as an example for volume. and h is the height. and two faces have areas of w h. two faces have areas of l h. each with the same length. and you would find Volume = (3x)(2x)(x) = 6x3. Here¶s the figure again: All you have to do is enter the given values into the formula for surface area: Diagonal Length of a Rectangular Solid Page 97 The diagonal of a rectangular solid. that connect each pair of opposite vertices. is the line segment whose endpoints are opposite corners of the solid. Notice how this formula corresponds with the general formula for the volume of a prism: the product lw is the area of the base. The six faces of a rectangular solid consist of three congruent pairs. l = 3x. and h = 2x. To practice. w is the width. Now try to find the volume of the prism in the following example: In this solid. Here¶s one diagonal drawn in: . w = x. w is the width.where l is the length. In other words. You can look at this formula as the Pythagorean theorem in three dimensions. This is . and DE) is . and height of the rectangular solid. incorporating height to find the length of . Then use the Pythagorean theorem again. you can derive this formula using the Pythagorean theorem. and height are all equal. Page The formula for finding the volume of a cube is essentially the same as the formula for the volume of a rectangular volume. the length. In fact. and height are equal. find the length of the diagonal along the base. However. CF. width. Thus. and CG = 3? The question gives the length. width. so you can just plug those numbers into the formula: The length of the diagonal AH (as well as BG. A Math IC question might ask you: What is the length of diagonal AH in the rectangular solid below if AC = 5. and h is the height. Cubes Just as a square is a special kind of rectangle. First. A cube is a rectangular solid whose edges are each the same length. the cube volume formula is: 98 Volume of a Cube . and each of its six faces is a square. w is the width.The formula for the length of a diagonal is: where l is the length. and the diagonal from one corner to the other: . width. a cube is a special kind of rectangular solid. GH = 6. since a cube¶s length. Using the volume formula: Surface Area of a Cylinder Page 99 . The height of a cylinder.where s is the length of one edge of the cube. The radius of a cylinder. with s = l = w = h: where s is the length of one edge of the cube. The formula for the diagonal of a cube is simply adapted from the formula for the diagonal length of a rectangular solid. Surface Area of a Cube Since a cube is just a rectangular solid whose sides are all equal. Volume of a Cylinder The volume of a cylinder is the product of the area of its base and its height. Because a cylinder has a circular base. is the length of the line segment whose endpoints are the centers of the bases. If you know the height and radius of a cylinder. Diagonal Length of a Cube The same is true for measuring the diagonal length of a cube. except with s = l = w = h: where s is the length of one edge of the cube. r. the volume of a cylinder is: where r is the radius of the circular base and h is the height. is the radius of its base. you can easily calculate its volume and surface area. h. the formula for finding the surface area of a cube is the same as that for a rectangular solid. Try to find the volume of the cylinder below: This cylinder has a radius of 4 and a height of 6. Cylinders A cylinder is a prism with circular bases. The lateral height. As with prisms. but not too difficult. Cones A cone is not a prism.The surface area of a cylinder is the sum of the areas of the two bases and the lateral face of the cylinder. The bases are congruent circles. The radius of a cone is the radius of its one circular base. Volume of a Cone Since a cone is similar to a cylinder except that it is collapsed to a single point at one end. so they cannot be considered prisms. the surface area of a cylinder is given by this formula: where r is the radius and h is the height. To find the surface area of the cylinder in the practice example on volume. radius. but it is similar to a cylinder. Notice that the height. or slant height. you will always be able to find the third by using the Pythagorean theorem. finding the surface area involves plugging the height and radius of the base into the formula. h. these three measurements are denoted by r. the formula for the volume of a cone is a fraction of the formula for the volume of a cylinder: Page 100 . of a cone is the distance from a point on the edge of the base to the apex. The height of a cone is the distance from the center of the base to the apex (the point on top). so their areas can be found easily. respectively. The lateral face is the tubing that connects the two bases. This means that if you know the value for any two of these measurements. Therefore. A cone is essentially a cylinder in which one of the bases is collapsed into a single point at the center of the base. and l. As with finding the volume of a cylinder. and lateral height of a cone form a right triangle. just plug the values into the formula: Solids That Aren¶t Prisms Some of the solids that appear on the Math IC do not have two congruent bases that lie in parallel planes. When ³unrolled.´ the lateral base is simply a rectangle whose length is the circumference of the base and whose width is the height of the cylinder. In the figure above. The formulas for the volume and surface area of the non-prisms are a little more complex than those for the prisms. you need to know how to calculate the volume and surface area of these non-prisms. be careful not to find only the lateral surface area and then stop. Because the base is a circle. For practice.´ which. Students often forget the step of adding on the area of the circular base. depending on the shape of the cone. 101 . This is the formula: where r is the radius and l is the lateral height. find the volume of the cone pictured below: To answer this question. The area of the lateral surface is related to the circumference of the circle times the lateral height. it has an area of r2. and h = x . Practice by finding the total surface area of the cone pictured below: Pyramids Page The total surface area is equal to the area of the base plus the area of the lateral surface. just use the formula for the volume of a cone with the following values plugged in: r = x. l. l = 2x.where r is the radius and h is the height. a half-circle. or a ³Pacman´ shape. The total surface area is the sum of the base area and lateral surface area: When you are finding the surface area of a cone. The total surface area therefore equals x2 + 2x2 = 3 x2. The volume is: Surface Area of a Cone The surface area of a cone consists of the lateral surface area and the area of the base. The lateral surface area = x 2x. The lateral surface is the cone ³unrolled. The area of the base = x2. can be the shape of a triangle with a curved base. you can calculate the area of each face individually using techniques from plane geometry.A pyramid is like a cone. and we¶ve seen that B = 32 = 9. B = 32 = 9. If you come across one of those rare questions that covers the topic. The base is simply a square. and we can use the properties of a 30-60-90 triangle to find their areas: Page 102 . you need to add together the area of the base and the areas of the four sides. Practice by finding the surface area of the same pyramid in the figure below: To calculate the surface area. Try to find the volume of the pyramid below: The base is just a square with a side of 3. except that it has a polygon for a base. The shaded area in the figure above is the base. and the height is 3 /2. you should be able to recognize them and calculate their volume. and the height is the perpendicular distance from the apex of the pyramid to its base. Volume of a Pyramid The formula for calculating the volume of a pyramid is: where B is the area of the base and h is the height. and the total volume of the pyramid is: Surface Area of a Pyramid The surface area of a pyramid is rarely tested on the Math IC test. Though pyramids are not tested very often on the Math IC test. since the base of a pyramid is a square and the sides are triangles. Each side is an equilateral triangle. Comparing Dimensions The first way the Math IC will test your understanding of the relationship among the basic measurements of geometric solids is by giving you the length. surface area. Area = 1 /2 3 =9 3 /2 = 9 / 4. Essentially. the center of the sphere. r. What is the ratio of the radius of the sphere to the radius of the cylinder? Page 103 . The test gauges your understanding by asking you to calculate the lengths. For example. The Math IC will ask you about the relationship between these three properties. you need to have a good grasp of the formulas for each type of solid and be able to relate those formulas to one another algebraically. Surface Area. but also whether you understand those formulas. But in order to do the math. The math needed to answer comparingdimensions questions isn¶t that hard. or volume of different solids and asking you to compare their dimensions. and Volume The Math IC tests not only whether you¶ve memorized the formulas for the different geometric solids.For each triangle. If you know the radius of a sphere you can find both its volume and surface area. The surface area of a sphere is the same as the volume of a cylinder. a sphere is a 3-D circle. The main measurement of a sphere is its radius. the distance from the center to any point on the sphere. and volumes of various solids. The equation for the volume of a sphere is: The equation for the surface area of a sphere is: Relating Length. The Math IC includes two kinds of questions covering these relationships. The sum of the areas of the four triangles is 4 9 /4 The total surface area of the pyramid is 9 + 9 Spheres A sphere is the collection of points in three-dimensional space that are equidistant from a fixed point. surface areas. This question tells you that the surface area of a sphere and the volume a cylinder are equal. and a volume of 104 Example 3 . where s is the length of a side. Remember that this rule holds true only if all of a solid¶s dimensions increase in length by a given factor. Replace s with 2s. by how much do you increase the area of that square? If you understand the formula for the area of a square. and height). the rule holds true when just a side or the radius changes. What is the volume of a rectangular solid of dimensions x /2 y /2 z? Page A rectangular solid has dimensions x y z (these are its length. cylinder. surface area. This ratio is given by rs/rc. and then asking how this change will influence the other measurements. The question asks for the ratio between the radii of the sphere and the cylinder. The formula for the area of a square is A = s2. Example 2 If a sphere¶s radius is halved. Now you can solve the equation 4 rs2 = rc2 h for the ratio rs/rc. which is the same thing as decreasing by a factor of 8. by what factor does its volume decrease? The radius of the sphere is multiplied by a factor of 1»2 (or divided by a factor of 2). all of the length dimensions must change by the same factor. width. A sphere¶s surface area is 4 (rs)2. A cylinder¶s volume is (rc)2 h. and you see that the area of a square quadruples when the length of its sides double: (2s)2 = 4s2. and so its volume multiplies by the cube of that factor: (1»2)3 = 1»8. if the height of a cylinder doubles but the radius of the base triples²you will have to go back to the equation for the dimension you are trying to determine and calculate by hand. Example 1 If you double the length of the side of a square. and volume is by changing one of these measurements by a given factor. or other solid. but for a rectangular solid. a simple rule can help you find the answer: y If a solid¶s length is multiplied by a given factor. So for a cube or a sphere. If the dimensions of the object do not increase by a constant factor²for instance. and its volume is multiplied by the cube of that factor. then the solid¶s surface area is multiplied by the square of that factor. the volume of the sphere is multiplied by a factor of 1»8 (divided by 8). When the lengths of a solid in the question are increased by a single constant factor. where rc is the radius of the cylinder. where rs is the radius of the sphere. Therefore. 64. this question is simple. and h is its height. Therefore. Changing Measurements The second way the Math IC will test your understanding of the relationships among length. and a rectangular solid inscribed in a sphere. we need to first find the value of r.If this rectangular solid had dimensions that were all one-half as large as the dimensions of the solid whose volume is 64. an inscribed-solid question will present a figure of an inscribed solid and give you information about one of the solids. what is the volume of the cylinder? The formula for the volume of a cylinder is r2(h). But dimension z is not multiplied by 1»2 like x and y. The figures below are. a cylinder inscribed in a sphere. x»2 y»2 z = 1»4 xyz = 1»4 64 = 16. in either the top or bottom face of the cube. These questions do require an ability to visualize inscribed solids and an awareness of how certain line segments relate to both of the solids in a given figure. of the cylinder. a sphere inscribed in a cube. Here¶s an example: In the figure below. Page 105 . So in order to solve for the volume of the cylinder. For example. To answer a question like this one. then its volume would be (1»2)3 64 = 1»8 64 = 8. So. d. The question states that h = 5. The key step in this problem is to recognize that the diagonal of a face of the cube is also the diameter. but there is no value given for r. Most often. say a rectangular solid. with the edges of the two solids touching. If the length of the diagonal of the cube is 4 and the height of the cylinder is 5. you should use the volume formula for rectangular solids: Volume = l w h. It is given in the question that xyz = 64. Math IC questions that involve inscribed solids don¶t require any techniques other than those you¶ve already learned. and then be asked to find the volume of the other solid. you may be given the radius of a cylinder. from left to right. Inscribed Solids An inscribed solid is a solid placed inside another solid. you need to use the radius of the cylinder to find the dimensions of the other solid so that you can answer the question. To see this. draw a diagonal. a cube is inscribed in a cylinder. Using the figure as your guide. or twice the radius. In order to find this diagonal. the radius of the cylinder is 4 / 2 = 2 the formula for the volume of the cylinder. which is the hypotenuse in a 45-45-90 triangle. The question states that the diagonal of the cube is 4 that s = 4. This means that the diagonal along a single face of the cube is 4 where s is the so it follows (using the special Plug that into properties of a 45-45-90 triangle). the diagonal of a face of the cube is equal to the diameter of the cylinder. The better you know the rules of inscribed solids. here are the rules of inscribed solids that most commonly appear on the Math IC. and you get (2 )2 5 = 40 . Helpful Tips Math IC questions involving inscribed solids are much easier to solve when you know how the lines of different solids relate to one another. For instance. Cylinder Inscribed in a Sphere The diameter of the sphere is equal to the diagonal of the cylinder¶s height and diameter. since the formula for the diagonal of a cube is s length of an edge of the cube. The best way to explain how this type of problem works is to provide a sample question: Page 106 . or s. Sphere Inscribed in a Cylinder Both the cylinder and the sphere have the same diameter and radius. So without further ado. Solids Produced by Rotating Polygons Another type of Math IC question that you may come across involves a solid produced by the rotation of a polygon. the better you¶ll do on these questions. Sphere Inscribed in a Cube The diameter of the sphere is equal to the length of the cube¶s edge. the previous example showed that when a cube is inscribed in a cylinder. Therefore. We can find s from the diagonal of the cube (not to be confused with the diagonal of a face of the cube). we need the length of an edge of the cube. when rotated a specific way. The question asks you to figure out the surface area of the cone. produce which solids. a cone is formed. being the side opposite the 30º angle. If the hypotenuse is 2. Now plug both values of l and r into the surface area formula and then simplify: Common Rotations You don¶t need to learn any new techniques or formulas for problems that deal with rotating figures. must be 1. You can easily calculate the length of BC since the triangle is a 30-60-90 triangle. To solve the problem. The radius of the circle is equal to side BC of the triangle. then BC.What is the surface area of the geometric solid produced by the triangle below when it is rotated 360 degrees about the axis AB? When this triangle is rotated about AB. you should see that the lateral height is equal to the hypotenuse of the triangle. Page 107 . Below is a summary of which polygons. the first thing you should do is sketch the cone that the triangle will form. The formula for surface area is r2 + rl. If you¶ve drawn your cone correctly. You just have to be able to visualize the rotation as it¶s described and be aware of which parts of the polygons become which parts of the geometric solid. which means you need to know the lateral height of the cone and the radius of the circle. . Surface Area of a Rectangular Solid Surface Area = 2lw + 2lh + 2wh. w is the width. where l is the length. and h is the height. Surface Area of a Cylinder Surface Area = 2 r2 + 2 rh. where s is the length of one edge. Surface Area of a Sphere Surface Area = 4 r2. where r is the radius and h is the height. A semicircle rotated about its diameter produces a sphere.A rectangle rotated about its edge produces a cylinder. Key Formulas Length of a Diagonal of a Cube d= d= =s where s is the length of one edge of the cube. where r is the radius of the base and l is the cone¶s lateral height. A right triangle rotated about one of its legs produces a cone. Total Surface Area of a Cone Page 108 . Length of a Diagonal of a Rectangular Solid Surface Area of a Cube Surface Area = 6s2. where r is the radius. A circle rotated about its diameter produces a sphere. A rectangle rotated about a central axis (which must contain the midpoints of both of the sides that it intersects) produces a cylinder. An isosceles triangle rotated about its axis of symmetry (the altitude from the vertex of the noncongruent angle) produces a cone. Lateral Surface Area of a Cone Lateral Surface Area = rl. and h is the height of the rectangular solid. where l is the length. w is the width. and h is the height. where r is the radius of the circular base and h is the cylinder¶s height. Volume of a Cube Volume = s3. where s is the length of one edge. the cone¶s surface area doubles. Volume of a Cone Volume = r2h. A Not enough information to tell rectangle stands so that its 6 inch side lies flat against the ground. If the length. and l is the cone¶s lateral height. where B is the area of the base and h is the height. where B is the area of the base. width. Review Questions 1. by what factor is the length of its diagonal multiplied? (A) (B) (C) (D) (D) 2. and height of a rectangular solid are all doubled. Volume of a Prism Volume = Bh. where r is the radius of the base. w is the width. If the rectangle is rotated Page 109 (D) 96 . Volume of a Sphere Volume = r3. (A) (B) (C) 48 2 4 8 A cylinder¶s radius is equal to its height. If its surface area is 100Ï . (A) (B) (C) (D) (D) 3. where r is the radius and h is the height.Surface Area = r2 + rl. When its radius and height are multiplied by the same factor. Volume of a Pyramid Volume = Bh. where r is the radius. and h is the height. what is its volume? 25Ï 50Ï 100Ï 125Ï 625Ï Cone A has volume 24. Volume of a Cylinder Volume = r2h. where l is the length. Volume of a Rectangular Solid Volume = lwh. What is Cone A¶s new volume? (D) 4. then its surface area multiplies by the square of that factor. its dimensions are multiplied by a factor of . the surface area of this cylinder is With the information given by the question. just like the rectangular solid¶s other dimensions. D . w. what is the volume of the resulting cylinder? (A) (B) (C) (D) (D) 24Ï 36Ï 64Ï 96Ï 144Ï Explanations 1. E Cone A¶s new volume is If the rectangle is rotated about a side of length 4. In the case of this cylinder. C where l. a. and 2h for these values: The length of the diagonal doubles. the volume becomes multiplied by the cube of that factor: Thus. 2w. and h are the dimensions of the solid. the radius is equal to height (r = h). the radius of the cylinder works out to 5. If the surface area of Cone A doubles. So if the dimensions of a cone are multiplied by the same factor. and its volume increases by the cube of that factor. Either way. 2.around the axis of one of its two 4 inch sides. to find r. If the dimensions of a cone are multiplied by the same factor. C The formula for a cone¶s surface area is r2 + rl. the cone¶s volume is multiplied by a factor of 4. which means that the volume of the cylinder is 53 = 125 . the solid¶s surface area is multiplied by the square of that factor. 3. so The volume of a cylinder is given by the formula Since we have a value only for the surface area of this cylinder. The formula for the length of the diagonal of a rectangular solid is Substitute 2l. we can find the radius by setting either 4 r2 = 100 or r2 = 100 »4 = 25. we must use the surface area formula. Page 110 In general. then the height of the cylinder will be 4 and the radius will be 6. A cone¶s volume is 1/ 3 r2h. for solids. which is Because r = h in this cylinder. if each dimension of a cone is multiplied by the same factor. Once you visualize the cylinder. Many of the basic concepts in this chapter may be familiar to you from plane geometry. and coordinate space. you can plug in the values for the volume of a cylinder: Coordinate Geometry COORDINATE GEOMETRY QUESTIONS make up about 10 percent of the Math IC test. such as slope. Page 111 . but they have a twist: the coordinate plane gives us new ways to analyze shapes and figures. Coordinate geometry also covers a number of topics that plane geometry doesn¶t. parabolas. 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